{"id":3922,"date":"2022-04-05T18:43:19","date_gmt":"2022-04-05T18:43:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=3922"},"modified":"2022-10-29T01:53:41","modified_gmt":"2022-10-29T01:53:41","slug":"differentiability","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/differentiability\/","title":{"raw":"Differentiability","rendered":"Differentiability"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Explain when a function of two variables is differentiable.<\/li>\r\n \t
  • Use the total differential to approximate the change in a function of two variables.<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<\/div>\r\nWhen working with a function [latex]y=f\\,(x)[\/latex] of one variable, the function is said to be differentiable at a point [latex]x=a[\/latex] if [latex]f'\\,(a)[\/latex] exists. Furthermore, if a function of one variable is differentiable at a point, the graph is \"smooth\" at that point (i.e., no corners exist) and a tangent line is well-defined at that point.\r\n\r\nThe idea\u00a0behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In this case, a surface is considered to be smooth at point [latex]P[\/latex]\u00a0if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at a point [latex](x_0,\\ y_0)[\/latex] is given by\r\n
    [latex]\\large{z=f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)}[\/latex],<\/div>\r\n \r\n\r\nFor a tangent plane to exist at the point [latex](x_0,\\ y_0)[\/latex], the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in Figure 3. In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin.\r\n
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    Definition<\/h3>\r\n\r\n
    \r\n\r\nA function [latex]f\\,(x,\\ y)[\/latex] is\u00a0differentiable<\/strong> at a point [latex]P\\ (x_0,\\ y_0)[\/latex] if, for all points [latex](x,\\ y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex]P[\/latex], we can write\r\n
    [latex]\\large{f\\,(x,\\ y,\\ z)=f\\,(x_0,\\ y_0)+f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)+E\\,(x,\\ y)}[\/latex],<\/div>\r\n \r\n\r\nwhere the error term [latex]E[\/latex] satisfies\r\n
    [latex]\\large{\\displaystyle\\lim_{(x,\\ y)\\to(x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}=0}.[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\nThe last term in the Differentiable Function at a Point Equation is referred to as the error term<\/em><\/span> and it represents how closely the tangent plane comes to the surface in a small neighborhood ([latex]\\delta[\/latex] disk) of point [latex]P[\/latex]. For the function [latex]f[\/latex] to be differentiable at [latex]P[\/latex], the function must be smooth\u2014that is, the graph of [latex]f[\/latex] must be close to the tangent plane for points near [latex]P[\/latex].\r\n
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    Example: Demonstrating Differentiability<\/h3>\r\nShow that the function [latex]f\\,(x,\\ y)=2x^{2}-4y[\/latex] is differentiable at point [latex](2,-3)[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1467793733125\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793733125\"]\r\n

    First, we calculate [latex]f\\,(x_0,\\ y_0)[\/latex], [latex]f_x\\,(x_0,\\ y_0)[\/latex], and [latex]f_y\\,(x_0,\\ y_0)[\/latex] using [latex]x_0=2[\/latex] and [latex]y_0=-3[\/latex], then we use\u00a0the equation for a Differentiable Function at a Point:<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {f\\,(2,-3)} & =\\hfill & {2(2)^{2}-4(-3)=8+12=20}\\hfill \\\\ \\hfill {f_x\\,(2,-3)} & =\\hfill & {4(2)=8}\\hfill \\\\ \\hfill {f_y\\,(2,-3)} & =\\hfill & {-4.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nTherefore [latex]m_1=8[\/latex] and [latex]m_2 =-4[\/latex], and\u00a0the equation for a Differentiable Function at a Point\u00a0becomes\r\n
    [latex]\\begin{array}{ccc}\\hfill {f\\,(x,\\ y)} & =\\hfill & {f\\,(2,-3)+f_x\\,(2,-3)(x-2)+f_y\\,(2,-3)(y+3)+E\\,(x,\\ y)}\\hfill \\\\ \\hfill {2x^{2}-4y} & =\\hfill & {20+8(x-2)-4(y+3)+E\\,(x,\\ y)}\\hfill \\\\ \\hfill {2x^{2}-4y} & =\\hfill & {20+8x-16-4y-12+E\\,(x,\\ y)}\\hfill \\\\ \\hfill {2x^{2}-4y}& =\\hfill & {8x-4y-8+E\\,(x,\\ y)}\\hfill \\\\ \\hfill{E\\,(x,\\ y)}& =\\hfill & {2x^{2}-8x+8.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nNext, we calculate [latex]\\displaystyle\\lim_{(x,\\ y)\\to (x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}[\/latex]:\r\n
    [latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to (x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}\\frac{2x^{2}-8x+8}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}\\frac{2(x^{2}-4x+4)}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}\\frac{2(x-2)^{2}}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}}\\hfill \\\\ \\hfill & \\leq\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}\\frac{2((x-2)^{2}+(y+3)^{2})}{\\sqrt{(x-2)^{2}+(y+3)^{2}}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}2\\sqrt{(x-2)^{2}+(y+3)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nSince [latex]E\\,(x,\\ y)\\geq 0[\/latex] for any value of [latex]x[\/latex] or [latex]y[\/latex], the original limit must be equal to zero. Therefore, [latex]f\\,(x,\\ y)=2x^{2}-4y[\/latex] is differentiable at point [latex](2,-3)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    TRY IT<\/h3>\r\nShow that the function [latex]f\\,(x,\\ y)=3x-4y^{2}[\/latex] is differentiable at point [latex](-1,\\ 2)[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1467793633124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793633124\"]\r\n
    [latex]f\\,(-1,\\ 2)=-19[\/latex], [latex]f_x\\,(-1,\\ 2)=3[\/latex], [latex]f_y\\,(-1,\\ 2)=-16[\/latex], [latex]E\\,(x,\\ y)=-4(y-2)^{2}[\/latex].<\/div>\r\n \r\n
    [latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to (x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}}& =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4(y-2)^{2}}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill & \\leq\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4((x+1)^{2}+(y-2)^{2})}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}-4\\sqrt{(x+1)^{2}+(y-2)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/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