{"id":3926,"date":"2022-04-05T18:45:58","date_gmt":"2022-04-05T18:45:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3926"},"modified":"2022-10-29T02:04:34","modified_gmt":"2022-10-29T02:04:34","slug":"implicit-differentiation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/implicit-differentiation\/","title":{"raw":"Implicit Differentiation","rendered":"Implicit Differentiation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Perform implicit differentiation of a function of two or more variables.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793956132\">Recall from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/introduction-to-implicit-differentiation\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"3-8-implicit-differentiation\">Implicit Differentiation<\/a>\u00a0that\u00a0<span id=\"f10a2a32-9eca-48cf-9e84-10c18a890484_term189\" class=\"no-emphasis\" data-type=\"term\">implicit differentiation<\/span>\u00a0provides a method for finding [latex]dy\/dx[\/latex]\u00a0when [latex]y[\/latex]\u00a0is defined implicitly as a function of [latex]x[\/latex]. The method involves differentiating both sides of the equation defining the function with respect to [latex]x[\/latex], then solving for [latex]dy\/dx[\/latex]. Partial derivatives provide an alternative to this method.<\/p>\r\n<p id=\"fs-id1167793420259\">Consider the ellipse defined by the equation [latex]x^{2}+3y^{2}+4y-4=0[\/latex]\u00a0as follows.<\/p>\r\n\r\n[caption id=\"attachment_1250\" align=\"aligncenter\" width=\"331\"]<img class=\"size-full wp-image-1250\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222043\/4-5-4.jpeg\" alt=\"An ellipse with center near (0, \u20130.7), major axis horizontal and of length roughly 4.5, and minor axis of length roughly 3.\" width=\"331\" height=\"272\" \/> Figure 1. Graph of the ellipse defined by [latex]x^{2}+3y^{2}+4y-4=0[\/latex].[\/caption]This equation implicitly defines [latex]y[\/latex]\u00a0as a function of [latex]x[\/latex]. As such, we can find the derivative [latex]dy\/dx[\/latex]\u00a0using the method of implicit differentiation:\r\n\r\n[latex]\r\n\r\n\\hspace{8cm} \\begin{align}\r\n\r\n\\frac{d}{dx}(x^2+3y^2+4y-4)&amp;=\\frac{d}{dx}(0) \\\\\r\n\r\n2x+6y\\frac{dy}{dx}+4\\frac{dy}{dx}&amp;=0 \\\\\r\n\r\n(6y+4)\\frac{dy}{dx}&amp;=-2x \\\\\r\n\r\n\\frac{dy}{dx}&amp;=-\\frac{x}{3y+2}.\r\n\r\n\\end{align}[\/latex]\r\n\r\nWe can also define a function [latex]z=f(x, y)[\/latex]\u00a0by using the left-hand side of the equation defining the ellipse. Then [latex]f(x, y)=x^{2}+3y^{2}+4y-4[\/latex]. The ellipse [latex]x^{2}+3y^{2}+4y-4=0[\/latex]\u00a0can then be described by the equation [latex]f(x, y)=0[\/latex]. Using this function and the following theorem gives us an alternative approach to calculating\u00a0[latex]dy\/dx[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: implicit differentiation of a function of two or more variables<\/h3>\r\n\r\n<hr \/>\r\n\r\nSuppose the\u00a0function [latex]z=f(x, y)[\/latex] defines\u00a0function [latex]y[\/latex] implicitly as a function\u00a0function [latex]y=g(x)[\/latex] of [latex]x[\/latex] via the equation\u00a0[latex]z=f(x, y)=0[\/latex]. Then\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}}[\/latex]<\/p>\r\nprovided\u00a0[latex]f_y(x, y)\\neq 0[\/latex].\r\n\r\nIf the equation [latex]z=f(x, y, z)=0[\/latex] defines [latex]z[\/latex] implicitly as a differentiable function of [latex]x[\/latex] and [latex]y[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial z}{\\partial x}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial z}}[\/latex] and\u00a0[latex]\\large{\\frac{\\partial z}{\\partial y}=-\\frac{\\partial f\/\\partial y}{\\partial f\/\\partial z}}[\/latex]<\/p>\r\nas long as\u00a0[latex]f_z(x, y, z)\\neq 0[\/latex].\r\n\r\n<\/div>\r\nThe equation for Implicit Differentiation of a Function of Two or More Variables is a direct consequence of\u00a0the Chain Rule for Two Independent Variables. In particular, if we assume that [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex] via the equation [latex]f(x, y)=0[\/latex], we can apply the chain rule to find [latex]dy\/dx[\/latex]:\r\n\r\n[latex]\\hspace{8cm} \\begin{align}\r\n\r\n\\frac{d}{dx}f(x,y)&amp;=\\frac{d}{dx}(0) \\\\\r\n\r\n\\frac{\\partial f}{\\partial x}\\cdot\\frac{dx}{dx}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{dy}{dx}&amp;=0 \\\\\r\n\r\n\\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{dy}{dx}&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nSolving this equation for [latex]dy\/dx[\/latex] gives\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex].\u00a0The Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] and [latex]y[\/latex]\u00a0can be derived in a similar fashion.\r\n<p id=\"fs-id1167794125864\">Let\u2019s now return to the problem that we started before the previous theorem. Using\u00a0Implicit Differentiation of a Function of Two or More Variables\u00a0and the function [latex]f(x, y)=x^{2}+3y^{2}+4y-4[\/latex], we can obtain<\/p>\r\n[latex]\\hspace{10cm} \\begin{align}\r\n\r\n\\frac{\\partial f}{\\partial x}&amp;=2x \\\\\r\n\r\n\\frac{\\partial f}{\\partial y}&amp;=6y+4.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThen\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] gives\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}=-\\frac{2x}{6y+4}=-\\frac{x}{3y+2}},[\/latex]<\/p>\r\nwhich is the same result obtained by the earlier use of implicit differentiation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: implicit differentiation by partial derivatives<\/h3>\r\na. Calculate [latex]dy\/dx[\/latex] if [latex]y[\/latex] is implicitly as a function of [latex]x[\/latex] via the equation\u00a0[latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex].\u00a0What is the equation of the tangent line to the graph of this curve at point [latex](2, 1)[\/latex]?\r\n\r\nb. Calculate [latex]\\partial z\/\\partial x[\/latex] and\u00a0[latex]\\partial z\/\\partial y[\/latex], given [latex]x^2e^y-yze^x=0[\/latex].\r\n\r\n[reveal-answer q=\"035821644\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"035821644\"]\r\n\r\na. Set [latex]f(x,y)=3x^2-2xy+y^2+4x-6y-11=0[\/latex], then calculate\u00a0[latex]f_x[\/latex] and\u00a0[latex]f_y[\/latex]:\r\n\r\n[latex]f_x=6x-2y+4[\/latex]\r\n\r\n[latex]f_y=-2x+2y-6.[\/latex]\r\n\r\nThe derivative given by\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}=-\\frac{6x-2y+4}{-2x+2y-6}=-\\frac{3x-y+2}{x-y+3}}[\/latex]<\/p>\r\nThe slope of the tangent line at point\u00a0[latex](2, 1)[\/latex] is given by\r\n<p style=\"text-align: center;\">[latex]\\left.\\frac{dy}{dx}\\right|_{(x,y)=(2,1)}=\\frac{3(2)-1+2}{2-1+3}=\\frac74.[\/latex]<\/p>\r\nTo find the equation of the tangent line, we use the point-slope form (Figure 6):\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\ny-y_0&amp;=m(x-x_0) \\\\\r\n\r\ny-1&amp;=\\frac74(x-2) \\\\\r\n\r\ny&amp;=\\frac74x-\\frac72+1 \\\\\r\n\r\ny&amp;=\\frac74x-\\frac52.\r\n\r\n\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1251\" align=\"aligncenter\" width=\"492\"]<img class=\"size-full wp-image-1251\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222214\/4-5-5.jpeg\" alt=\"A rotated ellipse with equation 3x2 \u2013 2xy + y2 + 4x \u2013 6y \u2013 11 = 0 and with tangent at (2, 1). The equation for the tangent is given by y = 7\/4 x \u2013 5\/2. The ellipse\u2019s major axis is parallel to the tangent line.\" width=\"492\" height=\"497\" \/> Figure 2. Graph of the rotated ellipse defined by [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex].[\/caption]b. We have [latex]f(x,y,z)=x^2e^y-yze^x[\/latex]. Therefore,[latex]\\hspace{8cm}\\begin{align}\\frac{\\partial f}{\\partial x}&amp;=2xe^y-yze^x \\\\\\frac{\\partial f}{\\partial y}&amp;=x^2e^y-ze^x \\\\\\frac{\\partial f}{\\partial x}&amp;=-ye^x.\\end{align}[\/latex]Using\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] and [latex]y[\/latex],\r\n\r\n[latex]\\hspace{6cm}\\begin{alignat}{2}\r\n\r\n\\frac{\\partial z}{\\partial x}&amp;=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y} &amp;\\quad &amp;\\quad &amp;\\quad &amp;\\quad\\frac{\\partial z}{\\partial y}&amp;=-\\frac{\\partial f\/\\partial y}{\\partial f\/\\partial z} \\\\\r\n\r\n&amp;=-\\frac{2xe^y-yze^x}{-ye^x} &amp;\\quad &amp;\\quad \\text{and} &amp;\\quad &amp;\\quad &amp;=-\\frac{x^2e^y-ze^x}{-ye^x} \\\\\r\n\r\n&amp;=\\frac{2xe^y-yze^x}{ye^x} &amp;\\quad &amp;\\quad &amp;\\quad &amp;\\quad &amp;=\\frac{x^2e^y-ze^x}{ye^x}.\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind\u00a0[latex]dy\/dx[\/latex] if\u00a0[latex]y[\/latex] is defined implicitly as a function of\u00a0[latex]x[\/latex] by the equation\u00a0[latex]x^{2}+xy-y^{2}+7x-3y-26=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point\u00a0[latex](3, -2)[\/latex]?\r\n\r\n[reveal-answer q=\"995467213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995467213\"]\r\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\left.\\frac{2x+y+7}{2y-x+3}\\right|_{(3,-2)}=\\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\\frac{11}4[\/latex]<\/p>\r\nEquation of the tangent line:\u00a0[latex]y=-\\frac{11}4x+\\frac{25}4[\/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=8186162&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=YSpLF3QfzK0&amp;video_target=tpm-plugin-wh05pp33-YSpLF3QfzK0\" 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\/CP4.27_transcript.html\">transcript for \u201cCP 4.27\u201d here (opens in new window).<\/a><\/center>&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]6060[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Perform implicit differentiation of a function of two or more variables.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793956132\">Recall from\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/introduction-to-implicit-differentiation\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"3-8-implicit-differentiation\">Implicit Differentiation<\/a>\u00a0that\u00a0<span id=\"f10a2a32-9eca-48cf-9e84-10c18a890484_term189\" class=\"no-emphasis\" data-type=\"term\">implicit differentiation<\/span>\u00a0provides a method for finding [latex]dy\/dx[\/latex]\u00a0when [latex]y[\/latex]\u00a0is defined implicitly as a function of [latex]x[\/latex]. The method involves differentiating both sides of the equation defining the function with respect to [latex]x[\/latex], then solving for [latex]dy\/dx[\/latex]. Partial derivatives provide an alternative to this method.<\/p>\n<p id=\"fs-id1167793420259\">Consider the ellipse defined by the equation [latex]x^{2}+3y^{2}+4y-4=0[\/latex]\u00a0as follows.<\/p>\n<div id=\"attachment_1250\" style=\"width: 341px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1250\" class=\"size-full wp-image-1250\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222043\/4-5-4.jpeg\" alt=\"An ellipse with center near (0, \u20130.7), major axis horizontal and of length roughly 4.5, and minor axis of length roughly 3.\" width=\"331\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-1250\" class=\"wp-caption-text\">Figure 1. Graph of the ellipse defined by [latex]x^{2}+3y^{2}+4y-4=0[\/latex].<\/p>\n<\/div>\n<p>This equation implicitly defines [latex]y[\/latex]\u00a0as a function of [latex]x[\/latex]. As such, we can find the derivative [latex]dy\/dx[\/latex]\u00a0using the method of implicit differentiation:<\/p>\n<p>[latex]\\hspace{8cm} \\begin{align}    \\frac{d}{dx}(x^2+3y^2+4y-4)&=\\frac{d}{dx}(0) \\\\    2x+6y\\frac{dy}{dx}+4\\frac{dy}{dx}&=0 \\\\    (6y+4)\\frac{dy}{dx}&=-2x \\\\    \\frac{dy}{dx}&=-\\frac{x}{3y+2}.    \\end{align}[\/latex]<\/p>\n<p>We can also define a function [latex]z=f(x, y)[\/latex]\u00a0by using the left-hand side of the equation defining the ellipse. Then [latex]f(x, y)=x^{2}+3y^{2}+4y-4[\/latex]. The ellipse [latex]x^{2}+3y^{2}+4y-4=0[\/latex]\u00a0can then be described by the equation [latex]f(x, y)=0[\/latex]. Using this function and the following theorem gives us an alternative approach to calculating\u00a0[latex]dy\/dx[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: implicit differentiation of a function of two or more variables<\/h3>\n<hr \/>\n<p>Suppose the\u00a0function [latex]z=f(x, y)[\/latex] defines\u00a0function [latex]y[\/latex] implicitly as a function\u00a0function [latex]y=g(x)[\/latex] of [latex]x[\/latex] via the equation\u00a0[latex]z=f(x, y)=0[\/latex]. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}}[\/latex]<\/p>\n<p>provided\u00a0[latex]f_y(x, y)\\neq 0[\/latex].<\/p>\n<p>If the equation [latex]z=f(x, y, z)=0[\/latex] defines [latex]z[\/latex] implicitly as a differentiable function of [latex]x[\/latex] and [latex]y[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial z}{\\partial x}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial z}}[\/latex] and\u00a0[latex]\\large{\\frac{\\partial z}{\\partial y}=-\\frac{\\partial f\/\\partial y}{\\partial f\/\\partial z}}[\/latex]<\/p>\n<p>as long as\u00a0[latex]f_z(x, y, z)\\neq 0[\/latex].<\/p>\n<\/div>\n<p>The equation for Implicit Differentiation of a Function of Two or More Variables is a direct consequence of\u00a0the Chain Rule for Two Independent Variables. In particular, if we assume that [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex] via the equation [latex]f(x, y)=0[\/latex], we can apply the chain rule to find [latex]dy\/dx[\/latex]:<\/p>\n<p>[latex]\\hspace{8cm} \\begin{align}    \\frac{d}{dx}f(x,y)&=\\frac{d}{dx}(0) \\\\    \\frac{\\partial f}{\\partial x}\\cdot\\frac{dx}{dx}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{dy}{dx}&=0 \\\\    \\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{dy}{dx}&=0.    \\end{align}[\/latex]<\/p>\n<p>Solving this equation for [latex]dy\/dx[\/latex] gives\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex].\u00a0The Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] and [latex]y[\/latex]\u00a0can be derived in a similar fashion.<\/p>\n<p id=\"fs-id1167794125864\">Let\u2019s now return to the problem that we started before the previous theorem. Using\u00a0Implicit Differentiation of a Function of Two or More Variables\u00a0and the function [latex]f(x, y)=x^{2}+3y^{2}+4y-4[\/latex], we can obtain<\/p>\n<p>[latex]\\hspace{10cm} \\begin{align}    \\frac{\\partial f}{\\partial x}&=2x \\\\    \\frac{\\partial f}{\\partial y}&=6y+4.    \\end{align}[\/latex]<\/p>\n<p>Then\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] gives<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}=-\\frac{2x}{6y+4}=-\\frac{x}{3y+2}},[\/latex]<\/p>\n<p>which is the same result obtained by the earlier use of implicit differentiation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: implicit differentiation by partial derivatives<\/h3>\n<p>a. Calculate [latex]dy\/dx[\/latex] if [latex]y[\/latex] is implicitly as a function of [latex]x[\/latex] via the equation\u00a0[latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex].\u00a0What is the equation of the tangent line to the graph of this curve at point [latex](2, 1)[\/latex]?<\/p>\n<p>b. Calculate [latex]\\partial z\/\\partial x[\/latex] and\u00a0[latex]\\partial z\/\\partial y[\/latex], given [latex]x^2e^y-yze^x=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q035821644\">Show Solution<\/span><\/p>\n<div id=\"q035821644\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Set [latex]f(x,y)=3x^2-2xy+y^2+4x-6y-11=0[\/latex], then calculate\u00a0[latex]f_x[\/latex] and\u00a0[latex]f_y[\/latex]:<\/p>\n<p>[latex]f_x=6x-2y+4[\/latex]<\/p>\n<p>[latex]f_y=-2x+2y-6.[\/latex]<\/p>\n<p>The derivative given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dy}{dx}=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y}=-\\frac{6x-2y+4}{-2x+2y-6}=-\\frac{3x-y+2}{x-y+3}}[\/latex]<\/p>\n<p>The slope of the tangent line at point\u00a0[latex](2, 1)[\/latex] is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\left.\\frac{dy}{dx}\\right|_{(x,y)=(2,1)}=\\frac{3(2)-1+2}{2-1+3}=\\frac74.[\/latex]<\/p>\n<p>To find the equation of the tangent line, we use the point-slope form (Figure 6):<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    y-y_0&=m(x-x_0) \\\\    y-1&=\\frac74(x-2) \\\\    y&=\\frac74x-\\frac72+1 \\\\    y&=\\frac74x-\\frac52.    \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"attachment_1251\" style=\"width: 502px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1251\" class=\"size-full wp-image-1251\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222214\/4-5-5.jpeg\" alt=\"A rotated ellipse with equation 3x2 \u2013 2xy + y2 + 4x \u2013 6y \u2013 11 = 0 and with tangent at (2, 1). The equation for the tangent is given by y = 7\/4 x \u2013 5\/2. The ellipse\u2019s major axis is parallel to the tangent line.\" width=\"492\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-1251\" class=\"wp-caption-text\">Figure 2. Graph of the rotated ellipse defined by [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex].<\/p>\n<\/div>\n<p>b. We have [latex]f(x,y,z)=x^2e^y-yze^x[\/latex]. Therefore,[latex]\\hspace{8cm}\\begin{align}\\frac{\\partial f}{\\partial x}&=2xe^y-yze^x \\\\\\frac{\\partial f}{\\partial y}&=x^2e^y-ze^x \\\\\\frac{\\partial f}{\\partial x}&=-ye^x.\\end{align}[\/latex]Using\u00a0the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] and [latex]y[\/latex],<\/p>\n<p>[latex]\\hspace{6cm}\\begin{alignat}{2}    \\frac{\\partial z}{\\partial x}&=-\\frac{\\partial f\/\\partial x}{\\partial f\/\\partial y} &\\quad &\\quad &\\quad &\\quad\\frac{\\partial z}{\\partial y}&=-\\frac{\\partial f\/\\partial y}{\\partial f\/\\partial z} \\\\    &=-\\frac{2xe^y-yze^x}{-ye^x} &\\quad &\\quad \\text{and} &\\quad &\\quad &=-\\frac{x^2e^y-ze^x}{-ye^x} \\\\    &=\\frac{2xe^y-yze^x}{ye^x} &\\quad &\\quad &\\quad &\\quad &=\\frac{x^2e^y-ze^x}{ye^x}.    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find\u00a0[latex]dy\/dx[\/latex] if\u00a0[latex]y[\/latex] is defined implicitly as a function of\u00a0[latex]x[\/latex] by the equation\u00a0[latex]x^{2}+xy-y^{2}+7x-3y-26=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point\u00a0[latex](3, -2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q995467213\">Show Solution<\/span><\/p>\n<div id=\"q995467213\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\left.\\frac{2x+y+7}{2y-x+3}\\right|_{(3,-2)}=\\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\\frac{11}4[\/latex]<\/p>\n<p>Equation of the tangent line:\u00a0[latex]y=-\\frac{11}4x+\\frac{25}4[\/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=8186162&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=YSpLF3QfzK0&amp;video_target=tpm-plugin-wh05pp33-YSpLF3QfzK0\" 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\/CP4.27_transcript.html\">transcript for \u201cCP 4.27\u201d here (opens in new window).<\/a><\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm6060\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=6060&theme=oea&iframe_resize_id=ohm6060&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/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-3926\">\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 4.27. <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 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