Recall from Implicit Differentiation that implicit differentiation provides a method for finding [latex]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 defining the function with respect to [latex]x[/latex], then solving for [latex]dy/dx[/latex]. Partial derivatives provide an alternative to this method.

Consider the ellipse defined by the equation [latex]x^{2}+3y^{2}+4y-4=0[/latex] as follows.

Figure 1. Graph of the ellipse defined by [latex]x^{2}+3y^{2}+4y-4=0[/latex].

This equation implicitly defines [latex]y[/latex] as a function of [latex]x[/latex]. As such, we can find the derivative [latex]dy/dx[/latex] using the method of implicit differentiation:

[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]

We can also define a function [latex]z=f(x, y)[/latex] by 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] can 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 [latex]dy/dx[/latex].

The equation for Implicit Differentiation of a Function of Two or More Variables is a direct consequence of the 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]:

[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]

Solving this equation for [latex]dy/dx[/latex] gives the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[/latex]. The Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[/latex] and [latex]y[/latex] can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using Implicit Differentiation of a Function of Two or More Variables and the function [latex]f(x, y)=x^{2}+3y^{2}+4y-4[/latex], we can obtain

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

Then the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[/latex] gives

[latex]\large{\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}=-\frac{2x}{6y+4}=-\frac{x}{3y+2}},[/latex]

which is the same result obtained by the earlier use of implicit differentiation.

Example: implicit differentiation by partial derivatives

a. 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]?

b. Calculate [latex]\partial z/\partial x[/latex] and [latex]\partial z/\partial y[/latex], given [latex]x^2e^y-yze^x=0[/latex].

Show Solution

a. Set [latex]f(x,y)=3x^2-2xy+y^2+4x-6y-11=0[/latex], then calculate [latex]f_x[/latex] and [latex]f_y[/latex]:

[latex]f_x=6x-2y+4[/latex]

[latex]f_y=-2x+2y-6.[/latex]

The derivative given by

[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]

The slope of the tangent line at point [latex](2, 1)[/latex] is given by

[latex]\left.\frac{dy}{dx}\right|_{(x,y)=(2,1)}=\frac{3(2)-1+2}{2-1+3}=\frac74.[/latex]

To find the equation of the tangent line, we use the point-slope form (Figure 6):

[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]

 

Figure 2. Graph of the rotated ellipse defined by [latex]3x^{2}-2xy+y^{2}+4x-6y-11=0[/latex].

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 the Implicit Differentiation of a Function of Two or More Variables in terms of [latex]x[/latex] and [latex]y[/latex],

[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]

Watch the following video to see the worked solution to the above Try It