Recall from Implicit Differentiation that implicit differentiation provides a method for finding $dy/dx$ when $y$ is defined implicitly as a function of $x$. The method involves differentiating both sides of the equation defining the function with respect to $x$, then solving for $dy/dx$. Partial derivatives provide an alternative to this method.

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

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

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

$\hspace{8cm} \begin{align} \frac{d}{dx}(x^2+3y^2+4y-4)&=\frac{d}{dx}(0) \\ 2x+6y\frac{dy}{dx}+4\frac{dy}{dx}&=0 \\ (6y+4)\frac{dy}{dx}&=-2x \\ \frac{dy}{dx}&=-\frac{x}{3y+2}. \end{align}$

We can also define a function $z=f(x, y)$ by using the left-hand side of the equation defining the ellipse. Then $f(x, y)=x^{2}+3y^{2}+4y-4$. The ellipse $x^{2}+3y^{2}+4y-4=0$ can then be described by the equation $f(x, y)=0$. Using this function and the following theorem gives us an alternative approach to calculating $dy/dx$.

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 $y$ is defined implicitly as a function of $x$ via the equation $f(x, y)=0$, we can apply the chain rule to find $dy/dx$:

$\hspace{8cm} \begin{align} \frac{d}{dx}f(x,y)&=\frac{d}{dx}(0) \\ \frac{\partial f}{\partial x}\cdot\frac{dx}{dx}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}&=0 \\ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}&=0. \end{align}$

Solving this equation for $dy/dx$ gives the Implicit Differentiation of a Function of Two or More Variables in terms of $x$. The Implicit Differentiation of a Function of Two or More Variables in terms of $x$ and $y$ 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 $f(x, y)=x^{2}+3y^{2}+4y-4$, we can obtain

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

Then the Implicit Differentiation of a Function of Two or More Variables in terms of $x$ gives

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

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

Example: implicit differentiation by partial derivatives

a. Calculate $dy/dx$ if $y$ is implicitly as a function of $x$ via the equation $3x^{2}-2xy+y^{2}+4x-6y-11=0$. What is the equation of the tangent line to the graph of this curve at point $(2, 1)$?

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

Show Solution

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

$f_x=6x-2y+4$

$f_y=-2x+2y-6.$

The derivative given by

$\large{\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}=-\frac{6x-2y+4}{-2x+2y-6}=-\frac{3x-y+2}{x-y+3}}$

The slope of the tangent line at point $(2, 1)$ is given by

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

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

$\hspace{8cm}\begin{align} y-y_0&=m(x-x_0) \\ y-1&=\frac74(x-2) \\ y&=\frac74x-\frac72+1 \\ y&=\frac74x-\frac52. \end{align}$

 

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

b. We have $f(x,y,z)=x^2e^y-yze^x$. Therefore,$\hspace{8cm}\begin{align}\frac{\partial f}{\partial x}&=2xe^y-yze^x \\\frac{\partial f}{\partial y}&=x^2e^y-ze^x \\\frac{\partial f}{\partial x}&=-ye^x.\end{align}$Using the Implicit Differentiation of a Function of Two or More Variables in terms of $x$ and $y$,

$\hspace{6cm}\begin{alignat}{2} \frac{\partial z}{\partial x}&=-\frac{\partial f/\partial x}{\partial f/\partial y} &\quad &\quad &\quad &\quad\frac{\partial z}{\partial y}&=-\frac{\partial f/\partial y}{\partial f/\partial z} \\ &=-\frac{2xe^y-yze^x}{-ye^x} &\quad &\quad \text{and} &\quad &\quad &=-\frac{x^2e^y-ze^x}{-ye^x} \\ &=\frac{2xe^y-yze^x}{ye^x} &\quad &\quad &\quad &\quad &=\frac{x^2e^y-ze^x}{ye^x}. \end{alignat}$

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