Intro to Implicit Differentiat

Examples

The process of implicit differentiation is best understood by looking at examples. You will see that when we treat $y$ as a function of $x$ , often we must use the chain rule**.

Example: Find the equation of the tangent line to

$x^2+y^2=1$ at

$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .

Solution: First we find

$\frac{dy}{dx}$ so we can find the slope. We differentiate both sides of

$x^2+y^2=1$ with respect to

$x$ :

$\frac{d}{dx}\left(x^2+y^2\right) = \frac{d}{dx}(1).$ The derivative of

$x^2$ is

$2x$ . The derivative of the constant 1 is 0. By the chain rule,

$\frac{d}{dx}\left(y^2\right) = 2 y \cdot \frac{dy}{dx}^{**}.$ (**for example, say

$y=\frac{1}{x}$ . Then

$\frac{d}{dx}(y^2)=\frac{d}{dx}\left(\frac{1}{x}\right)^2=2\left(\frac{1}{x}\right)^1\frac{d}{dx}\left(\frac{1}{x}\right)=2 y \cdot \frac{dy}{dx}$ .) We now have

$\begin{eqnarray*}2x + 2y \frac{dy}{dx} &=& 0\\ 2y\frac{dy}{dx}&=&-2x\\ \frac{dy}{dx} &=& -\frac{x}{y}. \end{eqnarray*}$ We see that this answer involves both

$x$ and

$y$ , which is necessary to determine whether this derivative refers to the top of the circle or the bottom.

To figure out the slope at

$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ we just plug in these values to

$\frac{dy}{dx}$ to get the slope

$m= \frac{dy}{dx}\Big|_{\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)} = -\frac{\sqrt3/2}{1/2}=- \sqrt{3}.$ Therefore, the equation to the tangent line is

$y-\frac{1}{2}=-\sqrt 3\left(x-\frac{\sqrt 3}{2}\right)$ . DO: Write this equation in slope intercept form. Graph this tangent line along with the unit circle. Do the slope and

$y$ -intercept make sense? What would happen at the point

$\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ ?.

In the following video we set up the machinery of implicit differentiation and work more examples.