### 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.