{"id":1111,"date":"2025-10-15T14:19:02","date_gmt":"2025-10-15T14:19:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1111"},"modified":"2026-05-26T15:00:41","modified_gmt":"2026-05-26T15:00:41","slug":"the-chain-rule-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/the-chain-rule-learn-it-3\/","title":{"raw":"The Chain Rule: Learn It 3","rendered":"The Chain Rule: Learn It 3"},"content":{"raw":"

Implicit Differentiation<\/h2>\r\n

In single-variable calculus, we learned how to find [latex]\\frac{dy}{dx}[\/latex] when [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex]. The technique involved differentiating both sides of an equation with respect to [latex]x[\/latex], then solving for the derivative. Partial derivatives give us an alternative approach to this problem.<\/p>\r\n\r\n

\r\n

Implicit Differentiation for Single-Variable Functions<\/strong><\/p>\r\n

Implicit differentiation gives us a method for finding [latex]\\frac{dy}{dx}[\/latex] when [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex]. The method involves differentiating both sides of the equation with respect to [latex]x[\/latex], then solving for [latex]\\frac{dy}{dx}[\/latex].<\/p>\r\n\r\n<\/section>\r\n

Consider the ellipse defined by the equation [latex]x^{2}+3y^{2}+4y-4=0[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1250\" align=\"aligncenter\" width=\"331\"]\"An 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)&=\\frac{d}{dx}(0) \\\\\r\n\r\n2x+6y\\frac{dy}{dx}+4\\frac{dy}{dx}&=0 \\\\\r\n\r\n(6y+4)\\frac{dy}{dx}&=-2x \\\\\r\n\r\n\\frac{dy}{dx}&=-\\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] using the left-hand side of the equation. Then [latex]f(x,y)=x^{2}+3y^{2}+4y-4[\/latex], and the ellipse [latex]x^{2}+3y^{2}+4y-4=0[\/latex] can be described by [latex]f(x,y)=0[\/latex]. This gives us an alternative approach to calculating [latex]\\frac{dy}{dx}[\/latex].\r\n\r\n

\r\n

implicit differentiation of a function of two or more variables<\/h3>\r\n

Suppose [latex]z=f(x,y)[\/latex] defines [latex]y[\/latex] implicitly as a function [latex]y=g(x)[\/latex] via the equation [latex]f(x,y)=0[\/latex]. Then<\/p>\r\n

[latex]\\dfrac{dy}{dx}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial y}[\/latex]<\/p>\r\n

provided [latex]f_y(x,y)\\neq 0[\/latex].<\/p>\r\n

If the equation [latex]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>\r\n

[latex]\\dfrac{\\partial z}{\\partial x}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial z}[\/latex] and [latex]\\dfrac{\\partial z}{\\partial y}=-\\dfrac{\\partial f\/\\partial y}{\\partial f\/\\partial z}[\/latex]<\/p>\r\n

as long as [latex]f_z(x,y,z)\\neq 0[\/latex].<\/p>\r\n\r\n<\/section>This formula is a direct consequence of the chain rule. If [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex] via [latex]f(x,y)=0[\/latex], we can apply the chain rule to find [latex]\\frac{dy}{dx}[\/latex]:\r\n\r\n[latex]\\hspace{8cm} \\begin{align}\r\n\r\n\\frac{d}{dx}f(x,y)&=\\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}&=0 \\\\\r\n\r\n\\frac{\\partial f}{\\partial x}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{dy}{dx}&=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nSolving for [latex]\\frac{dy}{dx}[\/latex] gives us the formula. The formulas for three variables can be derived similarly.\r\n

Let's return to the ellipse example using this new approach. With [latex]f(x,y)=x^{2}+3y^{2}+4y-4[\/latex], we calculate:<\/p>\r\n[latex]\\hspace{10cm} \\begin{align}\r\n\r\n\\frac{\\partial f}{\\partial x}&=2x \\\\\r\n\r\n\\frac{\\partial f}{\\partial y}&=6y+4.\r\n\r\n\\end{align}[\/latex]\r\n

Then the formula gives:<\/p>\r\n

[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\n

which is the same result we obtained using the traditional implicit differentiation method.<\/p>\r\n\r\n

\r\n

Why Use Partial Derivatives? <\/strong>This method can be more efficient than traditional implicit differentiation, especially for complicated equations. Instead of differentiating the entire equation and solving for [latex]\\frac{dy}{dx}[\/latex], you just find two partial derivatives and plug them into the formula.<\/p>\r\n\r\n<\/section>

\r\n
    \r\n \t
  1. Calculate [latex]dy\/dx[\/latex] if [latex]y[\/latex] is implicitly as a function of [latex]x[\/latex] via the equation [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point [latex](2, 1)[\/latex]?<\/li>\r\n \t
  2. Calculate [latex]\\partial z\/\\partial x[\/latex] and\u00a0[latex]\\partial z\/\\partial y[\/latex], given [latex]x^2e^y-yze^x=0[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"035821644\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"035821644\"]\r\n
      \r\n \t
    1. 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]:[latex]f_x=6x-2y+4[\/latex][latex]f_y=-2x+2y-6.[\/latex]The derivative given by\r\n

      [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

      [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&=m(x-x_0) \\\\\r\n\r\ny-1&=\\frac74(x-2) \\\\\r\n\r\ny&=\\frac74x-\\frac72+1 \\\\\r\n\r\ny&=\\frac74x-\\frac52.\r\n\r\n\\end{align}[\/latex]\r\n\r\n \r\n\r\n[caption id=\"attachment_1251\" align=\"aligncenter\" width=\"492\"]\"A Figure 2. Graph of the rotated ellipse defined by [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex].[\/caption]<\/li>\r\n \t

    2. We have [latex]f(x,y,z)=x^2e^y-yze^x[\/latex]. Therefore,\r\n
      [latex]\\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]<\/center>\r\nUsing the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[\/latex] and [latex]y[\/latex]:
      [latex]\\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} \\\\<\/center>&=-\\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}.\r\n\r\n\\end{alignat}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]325843[\/ohm_question]<\/section>","rendered":"

      Implicit Differentiation<\/h2>\n

      In single-variable calculus, we learned how to find [latex]\\frac{dy}{dx}[\/latex] when [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex]. The technique involved differentiating both sides of an equation with respect to [latex]x[\/latex], then solving for the derivative. Partial derivatives give us an alternative approach to this problem.<\/p>\n

      \n

      Implicit Differentiation for Single-Variable Functions<\/strong><\/p>\n

      Implicit differentiation gives us a method for finding [latex]\\frac{dy}{dx}[\/latex] when [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex]. The method involves differentiating both sides of the equation with respect to [latex]x[\/latex], then solving for [latex]\\frac{dy}{dx}[\/latex].<\/p>\n<\/section>\n

      Consider the ellipse defined by the equation [latex]x^{2}+3y^{2}+4y-4=0[\/latex].<\/p>\n

      \"An
      Figure 1. Graph of the ellipse defined by [latex]x^{2}+3y^{2}+4y-4=0[\/latex].<\/figcaption><\/figure>\n

      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

      [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

      We can also define a function [latex]z=f(x,y)[\/latex] using the left-hand side of the equation. Then [latex]f(x,y)=x^{2}+3y^{2}+4y-4[\/latex], and the ellipse [latex]x^{2}+3y^{2}+4y-4=0[\/latex] can be described by [latex]f(x,y)=0[\/latex]. This gives us an alternative approach to calculating [latex]\\frac{dy}{dx}[\/latex].<\/p>\n

      \n

      implicit differentiation of a function of two or more variables<\/h3>\n

      Suppose [latex]z=f(x,y)[\/latex] defines [latex]y[\/latex] implicitly as a function [latex]y=g(x)[\/latex] via the equation [latex]f(x,y)=0[\/latex]. Then<\/p>\n

      [latex]\\dfrac{dy}{dx}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial y}[\/latex]<\/p>\n

      provided [latex]f_y(x,y)\\neq 0[\/latex].<\/p>\n

      If the equation [latex]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

      [latex]\\dfrac{\\partial z}{\\partial x}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial z}[\/latex] and [latex]\\dfrac{\\partial z}{\\partial y}=-\\dfrac{\\partial f\/\\partial y}{\\partial f\/\\partial z}[\/latex]<\/p>\n

      as long as [latex]f_z(x,y,z)\\neq 0[\/latex].<\/p>\n<\/section>\n

      This formula is a direct consequence of the chain rule. If [latex]y[\/latex] is defined implicitly as a function of [latex]x[\/latex] via [latex]f(x,y)=0[\/latex], we can apply the chain rule to find [latex]\\frac{dy}{dx}[\/latex]:<\/p>\n

      [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

      Solving for [latex]\\frac{dy}{dx}[\/latex] gives us the formula. The formulas for three variables can be derived similarly.<\/p>\n

      Let’s return to the ellipse example using this new approach. With [latex]f(x,y)=x^{2}+3y^{2}+4y-4[\/latex], we calculate:<\/p>\n

      [latex]\\hspace{10cm} \\begin{align} \\frac{\\partial f}{\\partial x}&=2x \\\\ \\frac{\\partial f}{\\partial y}&=6y+4. \\end{align}[\/latex]<\/p>\n

      Then the formula gives:<\/p>\n

      [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

      which is the same result we obtained using the traditional implicit differentiation method.<\/p>\n

      \n

      Why Use Partial Derivatives? <\/strong>This method can be more efficient than traditional implicit differentiation, especially for complicated equations. Instead of differentiating the entire equation and solving for [latex]\\frac{dy}{dx}[\/latex], you just find two partial derivatives and plug them into the formula.<\/p>\n<\/section>\n

      \n
        \n
      1. Calculate [latex]dy\/dx[\/latex] if [latex]y[\/latex] is implicitly as a function of [latex]x[\/latex] via the equation [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point [latex](2, 1)[\/latex]?<\/li>\n
      2. Calculate [latex]\\partial z\/\\partial x[\/latex] and\u00a0[latex]\\partial z\/\\partial y[\/latex], given [latex]x^2e^y-yze^x=0[\/latex].<\/li>\n<\/ol>\n