Introduction and Examples

In addition to taking derivatives of functions we can take derivatives of equations. After all, if both sides of an equation are always equal, then their rates of change are also equal! If the equation involves both $x$ and $y$, then the derivatives will involve $\frac{dy}{dx}$. We can then use algebra to solve for $\frac{dy}{dx}$.

Example: Find the slope of the line tangent to the circle $x^2+y^2=1$ at $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. Solution: We take the derivative of $x^2+y^2=1$: $$\frac{d}{dx}\left(x^2+y^2\right) = \frac{d}{dx}(1).$$ The derivative of $x^2$ is $2x$. The derivative of the constant 1 is 0. By the chain rule, $$\frac{d}{dx}\left(y^2\right) = 2 y \cdot \frac{dy}{dx}.$$ All together we get $$2x + 2y \frac{dy}{dx} = 0,$$ so that $$\frac{dy}{dx} = -\frac{x}{y}.$$ Notice that that answer involves both $x$ and $y$! We could solve for $y$ in terms of $x$ (getting $y = \pm \sqrt{1-x^2}$) to get an answer that just involves $x$, but that's not necessary. To figure out the slope at $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ we just plug in $x=\frac{\sqrt{3}}{2}$ and $y=\frac{1}{2}$ to get the slope $$m= \frac{dy}{dx}\Big|_{{x=\sqrt3/2},\, {y=1/2}} = -\frac{\sqrt3/2}{1/2}=- \sqrt{3}.$$

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