Worksheet for Section 3.7

Secton 3.7 is about implicit differentiation -- that is, how to find derivatives of functions that are not given as explicit formulas in the form $y=f(x)$, but rather as equations in the two variables $x$ and $y$, like the equation $x^2+y^2=1$ for the unit circle. An equation like this is said to define a function implicitly because there are, in fact, many functions that fit the equation in one way or another. When you take a derivative of an expression like $y^2$ with respect to the variable $x$, treat $y$ as a function of $x$ and use the Chain Rule from Section 3.4. Then, for example, $\frac{d}{dx}[y^2]=2y\frac{d}{dx}[y]=2yy'$. To find $y'$ for the unit circle equation $x^2+y^2=1$, take the derivative with respect to $x$ on both sides of the equation:

$$\frac{d}{dx}[x^2+y^2] =\frac{d}{dx}[1]$$

$$2x+2yy' =0$$

Now solve this for $y'$ -- your formula for $y'$ will still include the variable $y$, but that is a consequence of the implicit form used to describle $y$ in the first place. This technique for finding $y'$ is called implicit differentiation. Here are a few more examples -- in each case, find $y'$ as above by implicit differentiation:

$$xy=1\qquad x+y^2=2\qquad y^3+y^2-y+2x^2=3\qquad \sin y=x$$

Returning to the unit circle example for a moment, how can you use your formula for $y'$ to find an equation for the tangent line to the unit circle through the point $(\frac12,\frac{\sqrt3}2)$ on the circle? This technique can also be extended to finding second derivatives -- in the unit circle example, compute the second derivative $y''$ by taking another derivative with respect to $x$ in your formula for $y'$. Is the unit circle concave up or concave down at the point $(\frac12,\frac{\sqrt3}2)$?

One last application of this technique (which is not covered in the book, but is a nice applicaton of implicit differentiation) has to do with properties of logarithms, and how you can use them to simplify some otherwise horrible expressions to make derivatives easier to calculate. Consider the following equation:

$$y=\frac{(x+4)^4}{\sqrt{x^2+2}}$$

To simplify this, take logarithms on both sides, then use the properties of logarithms:

$$\ln(y) =\ln\left(\frac{(x+4)^4}{\sqrt{x^2+2}}\right)$$

$$=4\ln(x+4)-\frac12\ln(x^2+2)$$

Now use implicit differentiation to find $y'$:

$$\frac{d}{dx}[\ln(y)] =\frac{d}{dx}\left[4\ln(x+4)-\frac12\ln(x^2+2)\right]$$

$$\frac{y'}{y} =4\left(\frac1{x+4}\right)-\frac12\left(\frac{2x}{x^2+2}\right)$$

Now solve this for $y'$ -- in this case, you can plug the original formula for $y$ into your result to eliminate that variable from your formula for $y'$.