{"id":3916,"date":"2022-04-05T18:37:49","date_gmt":"2022-04-05T18:37:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3916"},"modified":"2022-10-29T01:19:54","modified_gmt":"2022-10-29T01:19:54","slug":"partial-derivatives","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/partial-derivatives\/","title":{"raw":"Partial Derivatives","rendered":"Partial Derivatives"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the partial derivatives of a function of two variables.<\/li>\r\n \t<li>Calculate the partial derivatives of a function of more than two variables.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Derivatives of a Function of Two Variables<\/h2>\r\nWhen studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of [latex]y[\/latex] as a function of [latex]x[\/latex]. Leibniz notation for the derivative is [latex]dy\/dx[\/latex], which implies that [latex]y[\/latex] is the dependent variable and [latex]x[\/latex] is the independent variable. For a function [latex]z=f\\,(x,\\ y)[\/latex] of two variables, [latex]x[\/latex] and [latex]y[\/latex] are the independent variables and [latex]z[\/latex] is the dependent variable. This raises two questions right away: How do we adapt\u00a0Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f\\,(x,\\ y)[\/latex] be a function fo two variables. Then the\u00a0<strong>partial derivative<\/strong> of [latex]f[\/latex] with respect to [latex]x[\/latex], written as [latex]\\partial f\/\\partial x[\/latex], or [latex]f_x[\/latex], is defined as\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=\\lim_{h\\to{0}}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex], written as [latex]\\partial f\/\\partial y[\/latex], or [latex]f_y[\/latex], is defined as\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=\\lim_{k\\to{0}}\\frac{f\\,(x,\\ y+k)-f\\,(x,\\ y)}{k}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\nThis definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the [latex]d[\/latex] in the original notation is replaced with the symbol [latex]\\partial[\/latex]. (This rounded \"[latex]d[\/latex]\" is usually called \"partial,\" so [latex]\\partial f\/\\partial x[\/latex] is spoken as the \"partial of [latex]f[\/latex] with respect to [latex]x[\/latex].\")\u00a0This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating Partial Derivatives from the Definition<\/h3>\r\nUse the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\n<p style=\"text-align: left;\">First, calculate [latex]f\\,(x+h,\\ y)[\/latex].<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y)} &amp; =\\hfill &amp; {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)+5y-12} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, substitute this into our definition and simplify:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{(x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{h}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12-x^{2}+3xy-2y^{2}+4x-5y+12}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{2xh+h^{2}-3hy-4h}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\frac{h(2x+h-3y-4)}{h}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\frac{\\partial f}{\\partial y}[\/latex], first calculate [latex]f\\,(x,\\ y+h)[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x,\\ y+h)} &amp; =\\hfill &amp; {x^{2}-3x(y+h)+2(y+h)^{2}-4x+5(y+h)-12}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, substitute this into our definition and simplify:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+h)-f\\,(x,\\ y)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{(x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12-x^{2}+3xy-2y^{2}+4x-5y+12}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{-3xk+4yk+2k^{2}+5k}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}\\frac{h(-3x+4y+2k+5)}{k}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{k\\to 0}(-3x+4y+2k+5)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nUse the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=4x^{2}+2xy-y^{2}+3x-2y+5.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1367793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367793933124\"]\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial x}=8x+2y+3,\\ \\frac{\\partial f}{\\partial y}=2x-2y-2[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix [latex]y[\/latex] and define [latex]g\\,(x)=f\\,(x,\\ y)[\/latex] as a function of [latex]x[\/latex]. Then\r\n<p style=\"text-align: center;\">[latex]g'\\,(x)=\\displaystyle\\lim_{h\\to 0}\\frac{g\\,(x+h)-g\\,(x)}{h}=\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}=\\frac{\\partial f}{\\partial x}.[\/latex]<\/p>\r\nThe same is true for calculating the partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex]. This time, fix [latex]x[\/latex] and define [latex]h\\,(y)=f\\,(x,\\ y)[\/latex] as a function of [latex]y.[\/latex] Then\r\n<p style=\"text-align: center;\">[latex]h'\\,(x)=\\displaystyle\\lim_{k\\to 0}\\frac{h\\,(x+h)-h\\,(x)}{k}=\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+k)-f\\,(x,\\ y)}{k}=\\frac{\\partial f}{\\partial y}.[\/latex]<\/p>\r\nAll differentiation rules from the Introduction to Derivatives apply.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating Partial Derivatives<\/h3>\r\nCalculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the following functions by holding the opposite variable constant then differentiating:\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex]f\\,(x,\\ y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/li>\r\n \t<li>[latex]g\\,(x,\\ y)=\\sin{(x^{2}y-2x+4)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1267793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1267793933114\"]\r\n<ol>\r\n \t<li style=\"text-align: left;\">To calculate [latex]\\partial f\/\\partial x[\/latex], treat the variable [latex]y[\/latex] as a constant. Then differentiate [latex]f\\,(x,\\ y)[\/latex] with respect to [latex]x[\/latex] using the sum, difference, and power rules:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[3xy\\big]+\\frac{\\partial}{\\partial x}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[4x\\big]+\\frac{\\partial}{\\partial x}\\big[5y\\big]-\\frac{\\partial}{\\partial x}\\big[12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y+0-4+0-0}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe\u00a0derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable [latex]x[\/latex], so they are treated as constant terms. The derivative of the second term is equal to the coefficient of [latex]x[\/latex], which is [latex]-3y[\/latex]. Calculating [latex]\\partial f\/\\partial y[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[3xy\\big]+\\frac{\\partial}{\\partial y}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[4x\\big]+\\frac{\\partial}{\\partial y}\\big[5y\\big]-\\frac{\\partial}{\\partial y}\\big[12\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y-0+5-0}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThese are the same answers obtained in the previous example<\/li>\r\n \t<li>To calculate [latex]\\partial g\/\\partial x[\/latex], create the variable [latex]y[\/latex] as a constant. Then differentiate [latex]g\\,(x,\\ y)[\/latex] with respect to [latex]x[\/latex] using the chain rule and power rule:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\frac{\\partial g}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\big[\\sin{(x^{2}y-2x+4)}\\big]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(2xy-2)\\cos{(x^{2}y-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial f\/\\partial y[\/latex], treat the variable [latex]x[\/latex] as a constant. Then differentiate [latex]g\\,(x,\\ y)[\/latex] with respect to [latex]y[\/latex] using the chain rule and power rule:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial g}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\big[\\sin{(x^{2}y-2x+4)}\\big]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}\\cos{(x^{2}-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nCalculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function [latex]f\\,(x,\\ y)=\\tan{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex] by holding the opposite variable constant, then differentiating.\r\n\r\n[reveal-answer q=\"fs-id1367703933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367703933124\"]\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial x}=(3x^{2}-6xy^{2})\\sec^{2}{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial y}=(-6x^{2}y+8y^{3})\\sec^{2}{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186153&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=LsVhLEj4oP8&amp;video_target=tpm-plugin-9m34t2ja-LsVhLEj4oP8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.13_transcript.html\">transcript for \u201cCP 4.13\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6041[\/ohm_question]\r\n\r\n<\/div>\r\nHow can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in [latex]\\mathbb{R}^{3}[\/latex]. If we remove the limit from the definition of the partial derivative with respect to [latex]x[\/latex], the difference quotient remains:\r\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/p>\r\nThis resembles the difference quotient for the derivative of a function of one variable, except for the presence of the [latex]y[\/latex] variable.\u00a0variable. Figure 1 illustrates a surface described by an arbitrary function [latex]z=f\\,(x,\\ y)[\/latex].\r\n\r\n[caption id=\"attachment_981\" align=\"aligncenter\" width=\"468\"]<img class=\"size-full wp-image-981\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225847\/4-3-1.jpeg\" alt=\"A complicated curve in xyz space with a secant line through the points (x, y, f(x, y)) and (x + h, y, f(x + h, y)).\" width=\"468\" height=\"395\" \/> Figure 1.\u00a0Secant line passing through the points [latex]\\small{(x,y,f(x,y))}[\/latex] and\u00a0[latex]\\small{(x+h,y,f(x+h,y))}[\/latex].[\/caption]In Figure 1, the value of [latex]h[\/latex] is positive. If we graph [latex]f\\,(x,\\ y)[\/latex] and [latex]f\\,(x+h,\\ y)[\/latex] for an arbitrary point [latex](x,\\ y)[\/latex], then the slope of the secant line passing through these two points is given by\r\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/p>\r\nThis line is parallel to the [latex]x[\/latex]-axis. Therefore, the slope of the secant line represents an average rate of change of the function [latex]f[\/latex] as we travel parallel to the [latex]x[\/latex]-axis. As [latex]h[\/latex] approaches zero, the slope of the secant line approaches the slope of the tangent line.\r\n\r\nIf we chose to change [latex]y[\/latex] instead of [latex]x[\/latex] by the same incremental value [latex]h[\/latex], then the secant line is parallel to the [latex]y[\/latex]-axisw and so is the tangent line. Therefore, [latex]\\partial f\/\\partial x[\/latex] represents the slope of the tangent line passing through the point [latex](x,\\ y,\\ f\\,(x,\\ y))[\/latex] parallel to the [latex]x[\/latex]-axis and [latex]\\partial f\/\\partial y[\/latex] represents the slope of the tangent line passing through the point\u00a0[latex](x,\\ y,\\ f\\,(x,\\ y))[\/latex] parallel to the [latex]y[\/latex]-axis. If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called\u00a0<i>directional derivatives<\/i>, which we discuss in Directional Derivatives and the Gradient.\r\n\r\nWe now return to the idea of contour maps, which we introduced in Functions of Several Variables. We can use a contour map to estimate partial derivatives of a function [latex]g\\,(x,\\ y)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Partial DErivatives from a Contour Map<\/h3>\r\nUse a contour map to estimate [latex]\\partial g\/\\partial x[\/latex] at the point [latex](\\sqrt{5},\\ 0)[\/latex] for the function\r\n<div style=\"text-align: center;\">[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/div>\r\n<div>[reveal-answer q=\"fs-id1167793943114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793943114\"]\r\n<p style=\"text-align: left;\">The following graph represents a contour map for the function [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/p>\r\n\r\n<div>[caption id=\"attachment_983\" align=\"aligncenter\" width=\"344\"]<img class=\"size-full wp-image-983\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28230034\/4-3-2.jpeg\" alt=\"A series of concentric circles with the center the origin. The first is marked c = 0 and has radius 3; the second is marked c = 1 and has radius slightly less than 3; and the third is marked c = 2 and has radius slightly more than 2. The graph is marked with the equation g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2).\" width=\"344\" height=\"347\" \/> Figure 2.\u00a0Contour map for the function\u00a0[latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex], using [latex]\\small{c=0,1,2}[\/latex], and [latex]\\small{3}[\/latex] ([latex]\\small{c=3}[\/latex] corresponds to the origin).[\/caption]<\/div>\r\n&nbsp;\r\n\r\nThe inner circle on the contour map corresponds to [latex]c=2[\/latex] and the next circle out corresponds to [latex]c=1[\/latex]. The first circle is given by the equation [latex]2=\\sqrt{9-x^{2}-y^{2}}[\/latex]; the second circle is given by the equation [latex]1=\\sqrt{9-x^{2}-y^{2}}[\/latex]. The first equation simplifies to [latex]x^{2}+y^{2}=5[\/latex] and the second equation simplifies to [latex]x^{2}+y^{2}=8[\/latex]. The [latex]x[\/latex]-intercept of the first circle is [latex](\\sqrt{5},\\ 0)[\/latex] and the [latex]x[\/latex]-intercept of the second circle is [latex](2\\sqrt{2},\\ 0)[\/latex]. We can estimate the value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5},\\ 0)[\/latex] using the slope formula:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x,\\ y)=(\\sqrt{5},\\ 0)}\\approx\\frac{g\\,(\\sqrt{5},\\ 0)-g\\,(2\\sqrt{2},\\ 0)}{\\sqrt{5}-2\\sqrt{2}}=\\frac{2-1}{\\sqrt{5}-2\\sqrt{2}}=\\frac{1}{\\sqrt{5}-2\\sqrt{2}}\\approx -1.688.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate the exact value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5},\\ 0)[\/latex], we start by finding [latex]\\partial g\/\\partial x[\/latex] using the chain rule. First, we rewrite the function as [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}=(9-x^{2}-y^{2})^{\\frac{1}{2}}[\/latex] and then differentiate with respect to [latex]x[\/latex] while holding [latex]y[\/latex] constant:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}=\\frac{1}{2}(9-x^{2}-y^{2})^{-\\frac{1}{2}}(-2x)=-\\frac{x}{\\sqrt{9-x^{2}-y^{2}}}.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext, we evaluate this expression using [latex]x=\\sqrt{5}[\/latex] and [latex]y=0[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x,\\ y)=(\\sqrt{5},\\ 0)}=-\\frac{\\sqrt{5}}{\\sqrt{9-(\\sqrt{5})^{2}-(0)^{2}}}=-\\frac{\\sqrt{5}}{\\sqrt{4}}=-\\frac{\\sqrt{5}}{2}\\approx -1.118.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe estimate for the partial derivative corresponds to the slope of the secant line passing through the points [latex](\\sqrt{5},\\ 0,\\ g\\,(\\sqrt{5},\\ 0))[\/latex] and [latex](2\\sqrt{2},\\ 0,\\ g\\,(2\\sqrt{2},\\ 0))[\/latex]. It represents an approximation to the slope of the tangent line to the surface through the point [latex](\\sqrt{5},\\ 0,\\ g\\,(\\sqrt{5},\\ 0))[\/latex], which is parallel to the [latex]x[\/latex]-axis.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nUse a contour map to estimate [latex]\\partial f\/\\partial y[\/latex] at point [latex](0,\\ \\sqrt{2})[\/latex] for the function\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=x^{2}-y^{2}[\/latex].<\/p>\r\nCompare this with the exact answer.\r\n\r\n[reveal-answer q=\"fs-id1367703938124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367703938124\"]\r\n<p style=\"text-align: left;\">Using the curves corresponding to [latex]c=-2[\/latex] and [latex]c=-3[\/latex], we obtain<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}\\bigg|_{(x,\\ y)=(0,\\ \\sqrt{2})}\\approx\\frac{f\\,(0,\\ \\sqrt{3})-f\\,(0,\\ \\sqrt{2})}{\\sqrt{3}-\\sqrt{2}}=\\frac{-3-(-2)}{\\sqrt{3}-\\sqrt{2}}\\cdot\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}=-\\sqrt{3}-\\sqrt{2}\\approx-3.146}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe exact answer is\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}\\bigg|_{(x,\\ y)(0,\\ \\sqrt{2}}=-2y\\big|_{(x,\\ y)=(0,\\ \\sqrt{2})}=-2\\sqrt{2}\\approx -2.828.}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Functions of More Than Two Variables<\/h2>\r\nSuppose we have a function of three variables, such as [latex]w=f\\,(x,\\ y,\\ z)[\/latex]. We can calculate partial derivatives of [latex]w[\/latex] with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f\\,(x,\\ y,\\ z)[\/latex] be a function of three variables. Then, the\u00a0<em>partial derivative<\/em> of [latex]f[\/latex] with respect to [latex]x[\/latex], written as [latex]\\partial f\/\\partial x[\/latex], or [latex]f_x[\/latex], is defined to be\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y,\\ z)-f\\,(x,\\ y,\\ z)}{h}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe\u00a0<em>partial derivative<\/em> of [latex]f[\/latex] with respect to [latex]y[\/latex], written as [latex]\\partial f\/\\partial y[\/latex], or [latex]f_y[\/latex], is defined to be\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+k,\\ z)-f\\,(x,\\ y,\\ z)}{k}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThe\u00a0<em>partial derivative\u00a0<\/em>of [latex]f[\/latex] with respect to [latex]z[\/latex], written as [latex]\\partial f\/\\partial z[\/latex], or [latex]f_z[\/latex], is defined to be\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial z}=\\displaystyle\\lim_{m\\to 0}\\frac{f\\,(x,\\ y,\\ z+m)-f\\,(x,\\ y,\\ z)}{m}}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\nWe can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. Fore example, if we have a function [latex]f[\/latex] of [latex]x,\\ y[\/latex], and [latex]z[\/latex], and we wish to calculate [latex]\\partial f\/\\partial x[\/latex], then we create the other two independent variables as if they are constants, then differentiate with respect to [latex]x[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating partial derivatives for a Function of Three Variables<\/h3>\r\nUse the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nThen, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.\r\n\r\n[reveal-answer q=\"fs-id1367903938124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367903938124\"]\r\n<p style=\"text-align: left;\">We first calculate [latex]\\partial f\/\\partial x[\/latex] using our definition, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find [latex]\\partial f\/\\partial x[\/latex], we first need to calculate [latex]f\\,(x+h,\\ y,\\ z)[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y,\\ z)} &amp; =\\hfill &amp; {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)z+5yz^{2}-12(x+h)+4y-3z}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nand recall that [latex]f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z[\/latex]. Next, we substitute these two expressions into the equation:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z}{h} \\\\ - \\frac{x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{2xh+h^{2}-3hy-4hz-12h}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{h(2x+h-3y-4z-12)}{h}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4z-12)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2x-3y-4z-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThen we find [latex]\\partial f\/\\partial y[\/latex] by holding [latex]x[\/latex] and [latex]z[\/latex] constant. Therefore, any term that does not include the variable [latex]y[\/latex] is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:\r\n\r\n[latex] \\begin{alignat}{2} &amp;\\frac{\\partial}{\\partial y}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\\r\n\r\n\\hspace{2cm}&amp;=\\frac{\\partial}{\\partial y}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[3xy\\right]+\\frac{\\partial}{\\partial y}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[4xz\\right]+\\frac{\\partial}{\\partial y}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[12x\\right]+\\frac{\\partial}{\\partial y}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &amp;\\quad\\\\\r\n\r\n&amp;=0-3x+4y-0+5z^2-0+4-0\\\\\r\n\r\n&amp;=-3x+4y+5z^2+4.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\nTo calculate\u00a0[latex]\\partial f\/\\partial z[\/latex], we hold [latex]x[\/latex] and [latex]y[\/latex] constant and apply the sum, difference, and power rules for functions of one variable:\r\n\r\n[latex] \\begin{alignat}{2} &amp;\\frac{\\partial}{\\partial z}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\\r\n\r\n\\hspace{2cm}&amp;=\\frac{\\partial}{\\partial z}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[3xy\\right]+\\frac{\\partial}{\\partial z}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[4xz\\right]+\\frac{\\partial}{\\partial z}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[12x\\right]+\\frac{\\partial}{\\partial z}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &amp;\\quad\\\\\r\n\r\n&amp;=0-0+0-4x+10yz-0+0-3\\\\\r\n\r\n&amp;=-4z+10yz-3.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n&nbsp;\r\n<div style=\"text-align: center;\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nUse the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y,\\ z)=2x^{2}-4x^{2}+2y^{2}+5xz^{2}-6x+3z-8[\/latex].<\/p>\r\nThen, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.\r\n\r\n[reveal-answer q=\"fs-id1367703930124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367703930124\"]\r\n\r\n[latex] \\begin{alignat}{2} \\hspace{5cm} \\frac{\\partial f}{\\partial x}&amp;=4x-8xy+5z^{2}-6 \\\\\r\n\r\n\\frac{\\partial f}{\\partial y}&amp; = -4x^{2}+4y&amp;\\quad\\\\\r\n\r\n\\frac{\\partial f}{\\partial z}&amp; =10xz+3\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n<div style=\"text-align: left;\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating Partial Derivatives for a Function of Three Variables<\/h3>\r\nCalculate the three partial derivatives of the following functions.\r\n<ol>\r\n \t<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{x^{2}y-4xz+y^{2}}{x-3yz}[\/latex]<\/li>\r\n \t<li>[latex]g\\,(x,\\ y,\\ z)=\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1367703960124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367703960124\"]\r\n<div style=\"text-align: left;\">In each case, treat all variables as constants except the one whose partial derivative you are calculating.<\/div>\r\n&nbsp;\r\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}1. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial x}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial x}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(2xy-4z)(x-3yz)-(x^{2}y-4xz+y^{2})(1)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{2x^{2}y-6xy^{2}z-4xz+12yz^{2}-x^{2}y+4xz-y^{2}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{2}y-6xy^{2}z-4xz+12yz^{2}+4xz-y^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial y}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial y}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(x^{2}+2y)(x-3yz)-(x^{2}y-4xz+y^{2})(-3z)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{3}-3x^{2}yz+2xy-6y^{2}z+3x^{2}yz-12xz^{2}+3y^{2}z}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{x^{3}+2xy-3y^{2}z-12xz^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial z}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\frac{\\partial}{\\partial z}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial z}(x-3yz)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{(-4x)(x-3yz)-(x^{2}y-4xz+y^{2})(-3y)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4x^{2}+12xyz+3x^{2}y^{2}-12xyz+3y^{3}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{-4x^{2}+3x^{2}y^{2}+3y^{3}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}2. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial x}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial x}(x^{2}-yz)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2xy\\cos{(x^{2}y-z)}-2x\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial y}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial y}(x^{2}-yz)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {x^{2}\\cos{(x^{2}y-z)}+z\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial z}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial z}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial z}(x^{2}-yz)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-\\cos{(x^{2}y-z)}+y\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nCalculate [latex]\\partial f\/\\partial x,\\ \\partial f\/\\partial y[\/latex], and [latex]\\partial f\/\\partial z[\/latex] for the function [latex]f\\,(x,\\ y,\\ z)=\\sec{(x^{2}y)}-\\tan{(x^{3}yz^{2}})[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1367773930124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367773930124\"]\r\n\r\n[latex] \\begin{alignat}{2} \\hspace{5cm} \\frac{\\partial f}{\\partial x}&amp;=2xy\\sec{(x^{2}y)}\\tan{(x^{2}y)}-3x^{2}yz^{2}\\sec^{2}{(x^{3}yz^{2})} \\\\\r\n\r\n\\frac{\\partial f}{\\partial y}&amp;=x^{2}\\sec{(x^{2}y)}\\tan{(x^{2}y)-x^{3}z^{2}}\\sec^{2}{(x^{3}yz^{2})}&amp;\\quad\\\\\r\n\r\n\\frac{\\partial f}{\\partial z}&amp;=-2x^{3}yz\\sec^{2}{(x^{3}yz^{2})}\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186154&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cp5oduxbhuA&amp;video_target=tpm-plugin-9qc3yru5-cp5oduxbhuA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.16_transcript.html\">transcript for \u201cCP 4.16\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the partial derivatives of a function of two variables.<\/li>\n<li>Calculate the partial derivatives of a function of more than two variables.<\/li>\n<\/ul>\n<\/div>\n<h2>Derivatives of a Function of Two Variables<\/h2>\n<p>When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of [latex]y[\/latex] as a function of [latex]x[\/latex]. Leibniz notation for the derivative is [latex]dy\/dx[\/latex], which implies that [latex]y[\/latex] is the dependent variable and [latex]x[\/latex] is the independent variable. For a function [latex]z=f\\,(x,\\ y)[\/latex] of two variables, [latex]x[\/latex] and [latex]y[\/latex] are the independent variables and [latex]z[\/latex] is the dependent variable. This raises two questions right away: How do we adapt\u00a0Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]f\\,(x,\\ y)[\/latex] be a function fo two variables. Then the\u00a0<strong>partial derivative<\/strong> of [latex]f[\/latex] with respect to [latex]x[\/latex], written as [latex]\\partial f\/\\partial x[\/latex], or [latex]f_x[\/latex], is defined as<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=\\lim_{h\\to{0}}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex], written as [latex]\\partial f\/\\partial y[\/latex], or [latex]f_y[\/latex], is defined as<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=\\lim_{k\\to{0}}\\frac{f\\,(x,\\ y+k)-f\\,(x,\\ y)}{k}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the [latex]d[\/latex] in the original notation is replaced with the symbol [latex]\\partial[\/latex]. (This rounded &#8220;[latex]d[\/latex]&#8221; is usually called &#8220;partial,&#8221; so [latex]\\partial f\/\\partial x[\/latex] is spoken as the &#8220;partial of [latex]f[\/latex] with respect to [latex]x[\/latex].&#8221;)\u00a0This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Calculating Partial Derivatives from the Definition<\/h3>\n<p>Use the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">First, calculate [latex]f\\,(x+h,\\ y)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y)} & =\\hfill & {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)+5y-12} \\hfill \\\\ \\hfill & =\\hfill & {x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, substitute this into our definition and simplify:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{(x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{h}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{x^{2}+2xh+h^{2}-3xy-3hy+2y^{2}-4x-4h+5y-12-x^{2}+3xy-2y^{2}+4x-5y+12}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{2xh+h^{2}-3hy-4h}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\frac{h(2x+h-3y-4)}{h}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4)}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\frac{\\partial f}{\\partial y}[\/latex], first calculate [latex]f\\,(x,\\ y+h)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x,\\ y+h)} & =\\hfill & {x^{2}-3x(y+h)+2(y+h)^{2}-4x+5(y+h)-12}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, substitute this into our definition and simplify:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+h)-f\\,(x,\\ y)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{(x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12)-(x^{2}-3xy+2y^{2}-4x+5y-12)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{x^{2}-3xy-3xh+2y^{2}+4yh+2h^{2}-4x+5y+5h-12-x^{2}+3xy-2y^{2}+4x-5y+12}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{-3xk+4yk+2k^{2}+5k}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}\\frac{h(-3x+4y+2k+5)}{k}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{k\\to 0}(-3x+4y+2k+5)}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Use the definition of the partial derivative as a limit to calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=4x^{2}+2xy-y^{2}+3x-2y+5.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367793933124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367793933124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial x}=8x+2y+3,\\ \\frac{\\partial f}{\\partial y}=2x-2y-2[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix [latex]y[\/latex] and define [latex]g\\,(x)=f\\,(x,\\ y)[\/latex] as a function of [latex]x[\/latex]. Then<\/p>\n<p style=\"text-align: center;\">[latex]g'\\,(x)=\\displaystyle\\lim_{h\\to 0}\\frac{g\\,(x+h)-g\\,(x)}{h}=\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}=\\frac{\\partial f}{\\partial x}.[\/latex]<\/p>\n<p>The same is true for calculating the partial derivative of [latex]f[\/latex] with respect to [latex]y[\/latex]. This time, fix [latex]x[\/latex] and define [latex]h\\,(y)=f\\,(x,\\ y)[\/latex] as a function of [latex]y.[\/latex] Then<\/p>\n<p style=\"text-align: center;\">[latex]h'\\,(x)=\\displaystyle\\lim_{k\\to 0}\\frac{h\\,(x+h)-h\\,(x)}{k}=\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+k)-f\\,(x,\\ y)}{k}=\\frac{\\partial f}{\\partial y}.[\/latex]<\/p>\n<p>All differentiation rules from the Introduction to Derivatives apply.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Calculating Partial Derivatives<\/h3>\n<p>Calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the following functions by holding the opposite variable constant then differentiating:<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]f\\,(x,\\ y)=x^{2}-3xy+2y^{2}-4x+5y-12[\/latex].<\/li>\n<li>[latex]g\\,(x,\\ y)=\\sin{(x^{2}y-2x+4)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1267793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1267793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li style=\"text-align: left;\">To calculate [latex]\\partial f\/\\partial x[\/latex], treat the variable [latex]y[\/latex] as a constant. Then differentiate [latex]f\\,(x,\\ y)[\/latex] with respect to [latex]x[\/latex] using the sum, difference, and power rules:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[3xy\\big]+\\frac{\\partial}{\\partial x}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial x}\\big[4x\\big]+\\frac{\\partial}{\\partial x}\\big[5y\\big]-\\frac{\\partial}{\\partial x}\\big[12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y+0-4+0-0}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The\u00a0derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable [latex]x[\/latex], so they are treated as constant terms. The derivative of the second term is equal to the coefficient of [latex]x[\/latex], which is [latex]-3y[\/latex]. Calculating [latex]\\partial f\/\\partial y[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[x^{2}-3xy+2y^{2}-4x+5y-12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[x^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[3xy\\big]+\\frac{\\partial}{\\partial y}\\big[2y^{2}\\big]-\\frac{\\partial}{\\partial y}\\big[4x\\big]+\\frac{\\partial}{\\partial y}\\big[5y\\big]-\\frac{\\partial}{\\partial y}\\big[12\\big]}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y-0+5-0}\\hfill \\\\ \\hfill & =\\hfill & {-3x+4y+5.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>These are the same answers obtained in the previous example<\/li>\n<li>To calculate [latex]\\partial g\/\\partial x[\/latex], create the variable [latex]y[\/latex] as a constant. Then differentiate [latex]g\\,(x,\\ y)[\/latex] with respect to [latex]x[\/latex] using the chain rule and power rule:\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\frac{\\partial g}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\big[\\sin{(x^{2}y-2x+4)}\\big]} \\hfill \\\\ \\hfill & =\\hfill & {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]}\\hfill \\\\ \\hfill & =\\hfill & {(2xy-2)\\cos{(x^{2}y-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial f\/\\partial y[\/latex], treat the variable [latex]x[\/latex] as a constant. Then differentiate [latex]g\\,(x,\\ y)[\/latex] with respect to [latex]y[\/latex] using the chain rule and power rule:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial g}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\big[\\sin{(x^{2}y-2x+4)}\\big]}\\hfill \\\\ \\hfill & =\\hfill & {\\cos{(x^{2}y-2x+4)}\\frac{\\partial}{\\partial x}[x^{2}y-2x+4]} \\hfill \\\\ \\hfill & =\\hfill & {x^{2}\\cos{(x^{2}-2x+4)}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Calculate [latex]\\partial f\/\\partial x[\/latex] and [latex]\\partial f\/\\partial y[\/latex] for the function [latex]f\\,(x,\\ y)=\\tan{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex] by holding the opposite variable constant, then differentiating.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367703933124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367703933124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial x}=(3x^{2}-6xy^{2})\\sec^{2}{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\partial f}{\\partial y}=(-6x^{2}y+8y^{3})\\sec^{2}{(x^{3}-3x^{2}y^{2}+2y^{4})}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186153&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=LsVhLEj4oP8&amp;video_target=tpm-plugin-9m34t2ja-LsVhLEj4oP8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.13_transcript.html\">transcript for \u201cCP 4.13\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6041\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6041&theme=oea&iframe_resize_id=ohm6041&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in [latex]\\mathbb{R}^{3}[\/latex]. If we remove the limit from the definition of the partial derivative with respect to [latex]x[\/latex], the difference quotient remains:<\/p>\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/p>\n<p>This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the [latex]y[\/latex] variable.\u00a0variable. Figure 1 illustrates a surface described by an arbitrary function [latex]z=f\\,(x,\\ y)[\/latex].<\/p>\n<div id=\"attachment_981\" style=\"width: 478px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-981\" class=\"size-full wp-image-981\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225847\/4-3-1.jpeg\" alt=\"A complicated curve in xyz space with a secant line through the points (x, y, f(x, y)) and (x + h, y, f(x + h, y)).\" width=\"468\" height=\"395\" \/><\/p>\n<p id=\"caption-attachment-981\" class=\"wp-caption-text\">Figure 1.\u00a0Secant line passing through the points [latex]\\small{(x,y,f(x,y))}[\/latex] and\u00a0[latex]\\small{(x+h,y,f(x+h,y))}[\/latex].<\/p>\n<\/div>\n<p>In Figure 1, the value of [latex]h[\/latex] is positive. If we graph [latex]f\\,(x,\\ y)[\/latex] and [latex]f\\,(x+h,\\ y)[\/latex] for an arbitrary point [latex](x,\\ y)[\/latex], then the slope of the secant line passing through these two points is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\LARGE{\\frac{f\\,(x+h,\\ y)-f\\,(x,\\ y)}{h}}[\/latex].<\/p>\n<p>This line is parallel to the [latex]x[\/latex]-axis. Therefore, the slope of the secant line represents an average rate of change of the function [latex]f[\/latex] as we travel parallel to the [latex]x[\/latex]-axis. As [latex]h[\/latex] approaches zero, the slope of the secant line approaches the slope of the tangent line.<\/p>\n<p>If we chose to change [latex]y[\/latex] instead of [latex]x[\/latex] by the same incremental value [latex]h[\/latex], then the secant line is parallel to the [latex]y[\/latex]-axisw and so is the tangent line. Therefore, [latex]\\partial f\/\\partial x[\/latex] represents the slope of the tangent line passing through the point [latex](x,\\ y,\\ f\\,(x,\\ y))[\/latex] parallel to the [latex]x[\/latex]-axis and [latex]\\partial f\/\\partial y[\/latex] represents the slope of the tangent line passing through the point\u00a0[latex](x,\\ y,\\ f\\,(x,\\ y))[\/latex] parallel to the [latex]y[\/latex]-axis. If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called\u00a0<i>directional derivatives<\/i>, which we discuss in Directional Derivatives and the Gradient.<\/p>\n<p>We now return to the idea of contour maps, which we introduced in Functions of Several Variables. We can use a contour map to estimate partial derivatives of a function [latex]g\\,(x,\\ y)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Partial DErivatives from a Contour Map<\/h3>\n<p>Use a contour map to estimate [latex]\\partial g\/\\partial x[\/latex] at the point [latex](\\sqrt{5},\\ 0)[\/latex] for the function<\/p>\n<div style=\"text-align: center;\">[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793943114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793943114\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">The following graph represents a contour map for the function [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex].<\/p>\n<div>\n<div id=\"attachment_983\" style=\"width: 354px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-983\" class=\"size-full wp-image-983\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28230034\/4-3-2.jpeg\" alt=\"A series of concentric circles with the center the origin. The first is marked c = 0 and has radius 3; the second is marked c = 1 and has radius slightly less than 3; and the third is marked c = 2 and has radius slightly more than 2. The graph is marked with the equation g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2).\" width=\"344\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-983\" class=\"wp-caption-text\">Figure 2.\u00a0Contour map for the function\u00a0[latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex], using [latex]\\small{c=0,1,2}[\/latex], and [latex]\\small{3}[\/latex] ([latex]\\small{c=3}[\/latex] corresponds to the origin).<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The inner circle on the contour map corresponds to [latex]c=2[\/latex] and the next circle out corresponds to [latex]c=1[\/latex]. The first circle is given by the equation [latex]2=\\sqrt{9-x^{2}-y^{2}}[\/latex]; the second circle is given by the equation [latex]1=\\sqrt{9-x^{2}-y^{2}}[\/latex]. The first equation simplifies to [latex]x^{2}+y^{2}=5[\/latex] and the second equation simplifies to [latex]x^{2}+y^{2}=8[\/latex]. The [latex]x[\/latex]-intercept of the first circle is [latex](\\sqrt{5},\\ 0)[\/latex] and the [latex]x[\/latex]-intercept of the second circle is [latex](2\\sqrt{2},\\ 0)[\/latex]. We can estimate the value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5},\\ 0)[\/latex] using the slope formula:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x,\\ y)=(\\sqrt{5},\\ 0)}\\approx\\frac{g\\,(\\sqrt{5},\\ 0)-g\\,(2\\sqrt{2},\\ 0)}{\\sqrt{5}-2\\sqrt{2}}=\\frac{2-1}{\\sqrt{5}-2\\sqrt{2}}=\\frac{1}{\\sqrt{5}-2\\sqrt{2}}\\approx -1.688.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate the exact value of [latex]\\partial g\/\\partial x[\/latex] evaluated at the point [latex](\\sqrt{5},\\ 0)[\/latex], we start by finding [latex]\\partial g\/\\partial x[\/latex] using the chain rule. First, we rewrite the function as [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}=(9-x^{2}-y^{2})^{\\frac{1}{2}}[\/latex] and then differentiate with respect to [latex]x[\/latex] while holding [latex]y[\/latex] constant:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}=\\frac{1}{2}(9-x^{2}-y^{2})^{-\\frac{1}{2}}(-2x)=-\\frac{x}{\\sqrt{9-x^{2}-y^{2}}}.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next, we evaluate this expression using [latex]x=\\sqrt{5}[\/latex] and [latex]y=0[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial g}{\\partial x}\\bigg|_{(x,\\ y)=(\\sqrt{5},\\ 0)}=-\\frac{\\sqrt{5}}{\\sqrt{9-(\\sqrt{5})^{2}-(0)^{2}}}=-\\frac{\\sqrt{5}}{\\sqrt{4}}=-\\frac{\\sqrt{5}}{2}\\approx -1.118.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The estimate for the partial derivative corresponds to the slope of the secant line passing through the points [latex](\\sqrt{5},\\ 0,\\ g\\,(\\sqrt{5},\\ 0))[\/latex] and [latex](2\\sqrt{2},\\ 0,\\ g\\,(2\\sqrt{2},\\ 0))[\/latex]. It represents an approximation to the slope of the tangent line to the surface through the point [latex](\\sqrt{5},\\ 0,\\ g\\,(\\sqrt{5},\\ 0))[\/latex], which is parallel to the [latex]x[\/latex]-axis.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Use a contour map to estimate [latex]\\partial f\/\\partial y[\/latex] at point [latex](0,\\ \\sqrt{2})[\/latex] for the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=x^{2}-y^{2}[\/latex].<\/p>\n<p>Compare this with the exact answer.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367703938124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367703938124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">Using the curves corresponding to [latex]c=-2[\/latex] and [latex]c=-3[\/latex], we obtain<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}\\bigg|_{(x,\\ y)=(0,\\ \\sqrt{2})}\\approx\\frac{f\\,(0,\\ \\sqrt{3})-f\\,(0,\\ \\sqrt{2})}{\\sqrt{3}-\\sqrt{2}}=\\frac{-3-(-2)}{\\sqrt{3}-\\sqrt{2}}\\cdot\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}=-\\sqrt{3}-\\sqrt{2}\\approx-3.146}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The exact answer is<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}\\bigg|_{(x,\\ y)(0,\\ \\sqrt{2}}=-2y\\big|_{(x,\\ y)=(0,\\ \\sqrt{2})}=-2\\sqrt{2}\\approx -2.828.}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Functions of More Than Two Variables<\/h2>\n<p>Suppose we have a function of three variables, such as [latex]w=f\\,(x,\\ y,\\ z)[\/latex]. We can calculate partial derivatives of [latex]w[\/latex] with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]f\\,(x,\\ y,\\ z)[\/latex] be a function of three variables. Then, the\u00a0<em>partial derivative<\/em> of [latex]f[\/latex] with respect to [latex]x[\/latex], written as [latex]\\partial f\/\\partial x[\/latex], or [latex]f_x[\/latex], is defined to be<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=\\displaystyle\\lim_{h\\to 0}\\frac{f\\,(x+h,\\ y,\\ z)-f\\,(x,\\ y,\\ z)}{h}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The\u00a0<em>partial derivative<\/em> of [latex]f[\/latex] with respect to [latex]y[\/latex], written as [latex]\\partial f\/\\partial y[\/latex], or [latex]f_y[\/latex], is defined to be<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=\\displaystyle\\lim_{k\\to 0}\\frac{f\\,(x,\\ y+k,\\ z)-f\\,(x,\\ y,\\ z)}{k}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>The\u00a0<em>partial derivative\u00a0<\/em>of [latex]f[\/latex] with respect to [latex]z[\/latex], written as [latex]\\partial f\/\\partial z[\/latex], or [latex]f_z[\/latex], is defined to be<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial z}=\\displaystyle\\lim_{m\\to 0}\\frac{f\\,(x,\\ y,\\ z+m)-f\\,(x,\\ y,\\ z)}{m}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>We can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. Fore example, if we have a function [latex]f[\/latex] of [latex]x,\\ y[\/latex], and [latex]z[\/latex], and we wish to calculate [latex]\\partial f\/\\partial x[\/latex], then we create the other two independent variables as if they are constants, then differentiate with respect to [latex]x[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Calculating partial derivatives for a Function of Three Variables<\/h3>\n<p>Use the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Then, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367903938124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367903938124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">We first calculate [latex]\\partial f\/\\partial x[\/latex] using our definition, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find [latex]\\partial f\/\\partial x[\/latex], we first need to calculate [latex]f\\,(x+h,\\ y,\\ z)[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f\\,(x+h,\\ y,\\ z)} & =\\hfill & {(x+h)^{2}-3(x+h)y+2y^{2}-4(x+h)z+5yz^{2}-12(x+h)+4y-3z}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>and recall that [latex]f\\,(x,\\ y,\\ z)=x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z[\/latex]. Next, we substitute these two expressions into the equation:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{x^{2}+2xh+h^{2}-3xy-3xh+2y^{2}-4xz-4hz+5yz^{2}-12x-12h+4y-3z}{h} \\\\ - \\frac{x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{2xh+h^{2}-3hy-4hz-12h}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}\\left[\\frac{h(2x+h-3y-4z-12)}{h}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{h\\to 0}(2x+h-3y-4z-12)}\\hfill \\\\ \\hfill & =\\hfill & {2x-3y-4z-12.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Then we find [latex]\\partial f\/\\partial y[\/latex] by holding [latex]x[\/latex] and [latex]z[\/latex] constant. Therefore, any term that does not include the variable [latex]y[\/latex] is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:<\/p>\n<p>[latex]\\begin{alignat}{2} &\\frac{\\partial}{\\partial y}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\    \\hspace{2cm}&=\\frac{\\partial}{\\partial y}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[3xy\\right]+\\frac{\\partial}{\\partial y}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[4xz\\right]+\\frac{\\partial}{\\partial y}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial y}\\left[12x\\right]+\\frac{\\partial}{\\partial y}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &\\quad\\\\    &=0-3x+4y-0+5z^2-0+4-0\\\\    &=-3x+4y+5z^2+4.\\\\    \\end{alignat}[\/latex]<\/p>\n<p>To calculate\u00a0[latex]\\partial f\/\\partial z[\/latex], we hold [latex]x[\/latex] and [latex]y[\/latex] constant and apply the sum, difference, and power rules for functions of one variable:<\/p>\n<p>[latex]\\begin{alignat}{2} &\\frac{\\partial}{\\partial z}\\left[x^{2}-3xy+2y^{2}-4xz+5yz^{2}-12x+4y-3z\\right] \\\\    \\hspace{2cm}&=\\frac{\\partial}{\\partial z}\\left[x^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[3xy\\right]+\\frac{\\partial}{\\partial z}\\left[2y^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[4xz\\right]+\\frac{\\partial}{\\partial z}\\left[5yz^{2}\\right]-\\frac{\\partial}{\\partial z}\\left[12x\\right]+\\frac{\\partial}{\\partial z}\\left[4y\\right]-\\frac{\\partial}{\\partial y}\\left[3z\\right] &\\quad\\\\    &=0-0+0-4x+10yz-0+0-3\\\\    &=-4z+10yz-3.\\\\    \\end{alignat}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Use the limit definition of partial derivatives to calculate [latex]\\partial f\/\\partial x[\/latex] for the function<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y,\\ z)=2x^{2}-4x^{2}+2y^{2}+5xz^{2}-6x+3z-8[\/latex].<\/p>\n<p>Then, find [latex]\\partial f\/\\partial y[\/latex] and [latex]\\partial f\/\\partial z[\/latex] by setting the other two variables constant and differentiating accordingly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367703930124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367703930124\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{alignat}{2} \\hspace{5cm} \\frac{\\partial f}{\\partial x}&=4x-8xy+5z^{2}-6 \\\\    \\frac{\\partial f}{\\partial y}& = -4x^{2}+4y&\\quad\\\\    \\frac{\\partial f}{\\partial z}& =10xz+3\\\\    \\end{alignat}[\/latex]<\/p>\n<div style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Calculating Partial Derivatives for a Function of Three Variables<\/h3>\n<p>Calculate the three partial derivatives of the following functions.<\/p>\n<ol>\n<li>[latex]f\\,(x,\\ y,\\ z)=\\frac{x^{2}y-4xz+y^{2}}{x-3yz}[\/latex]<\/li>\n<li>[latex]g\\,(x,\\ y,\\ z)=\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367703960124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367703960124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: left;\">In each case, treat all variables as constants except the one whose partial derivative you are calculating.<\/div>\n<p>&nbsp;<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}1. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial x}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial x}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{(2xy-4z)(x-3yz)-(x^{2}y-4xz+y^{2})(1)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{2x^{2}y-6xy^{2}z-4xz+12yz^{2}-x^{2}y+4xz-y^{2}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{2}y-6xy^{2}z-4xz+12yz^{2}+4xz-y^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial y}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial y}(x-3yz)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{(x^{2}+2y)(x-3yz)-(x^{2}y-4xz+y^{2})(-3z)}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{3}-3x^{2}yz+2xy-6y^{2}z+3x^{2}yz-12xz^{2}+3y^{2}z}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{x^{3}+2xy-3y^{2}z-12xz^{2}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} & =\\hfill & {\\frac{\\partial}{\\partial z}\\bigg[\\frac{x^{2}y-4xz+y^{2}}{x-3yz}\\bigg]}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\frac{\\partial}{\\partial z}(x^{2}y-4xz+y^{2})(x-3yz)-(x^{2}y-4xz+y^{2})\\frac{\\partial}{\\partial z}(x-3yz)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{(-4x)(x-3yz)-(x^{2}y-4xz+y^{2})(-3y)}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4x^{2}+12xyz+3x^{2}y^{2}-12xyz+3y^{3}}{(x-3yz)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{-4x^{2}+3x^{2}y^{2}+3y^{3}}{(x-3yz)^{2}}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\begin{array}{ccc}2. \\hspace{1cm} {\\frac{\\partial f}{\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial x}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial x}(x^{2}-yz)}\\hfill \\\\ \\hfill & =\\hfill & {2xy\\cos{(x^{2}y-z)}-2x\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial y}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial y}(x^{2}-yz)}\\hfill \\\\ \\hfill & =\\hfill & {x^{2}\\cos{(x^{2}y-z)}+z\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill {\\frac{\\partial f}{\\partial z}} & =\\hfill & {\\frac{\\partial}{\\partial z}\\left[\\sin{(x^{2}y-z)}+\\cos{(x^{2}-yz)}\\right]}\\hfill \\\\ \\hfill & =\\hfill & {(\\cos{(x^{2}y-z)})\\frac{\\partial}{\\partial z}(x^{2}y-z)-(\\sin{(x^{2}-yz)})\\frac{\\partial}{\\partial z}(x^{2}-yz)} \\hfill \\\\ \\hfill & =\\hfill & {-\\cos{(x^{2}y-z)}+y\\sin{(x^{2}-yz)}} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Calculate [latex]\\partial f\/\\partial x,\\ \\partial f\/\\partial y[\/latex], and [latex]\\partial f\/\\partial z[\/latex] for the function [latex]f\\,(x,\\ y,\\ z)=\\sec{(x^{2}y)}-\\tan{(x^{3}yz^{2}})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367773930124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367773930124\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{alignat}{2} \\hspace{5cm} \\frac{\\partial f}{\\partial x}&=2xy\\sec{(x^{2}y)}\\tan{(x^{2}y)}-3x^{2}yz^{2}\\sec^{2}{(x^{3}yz^{2})} \\\\    \\frac{\\partial f}{\\partial y}&=x^{2}\\sec{(x^{2}y)}\\tan{(x^{2}y)-x^{3}z^{2}}\\sec^{2}{(x^{3}yz^{2})}&\\quad\\\\    \\frac{\\partial f}{\\partial z}&=-2x^{3}yz\\sec^{2}{(x^{3}yz^{2})}\\\\    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186154&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cp5oduxbhuA&amp;video_target=tpm-plugin-9qc3yru5-cp5oduxbhuA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.16_transcript.html\">transcript for \u201cCP 4.16\u201d here (opens in new window).<\/a><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3916\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.13. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 4.16. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 4.13\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 4.16\",\"author\":\"Ryan 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