Active Calculus 2nd Ed

In all of our studies with derivatives so far, we have worked with functions whose formula is given explicitly in terms of $x\text{.}$ But there are many interesting curves whose equations involving $x$ and $y$ are impossible to solve for $y$ in terms of $x\text{.}$

Perhaps the simplest and most natural of all such curves are circles. Because of the circle’s symmetry, for each $x$ value strictly between the endpoints of the horizontal diameter, there are two corresponding $y$-values. For instance, in Figure 2.7.1, we have labeled $A=(-3,\sqrt{7})$ and $B=(-3,-\sqrt{7})\text{,}$ and these points demonstrate that the circle fails the vertical line test. Hence, it is impossible to represent the circle through a single function of the form $y=f(x)\text{.}$ But portions of the circle can be represented explicitly as a function of $x\text{,}$ such as the highlighted arc that is magnified in the center of Figure 2.7.1. Moreover, it is evident that the circle is locally linear, so we ought to be able to find a tangent line to the curve at every point. Thus, it makes sense to wonder if we can compute $\frac{dy}{dx}$ at any point on the circle, even though we cannot write $y$ explicitly as a function of $x\text{.}$

We say that the equation $x^2+y^2=16$ defines $y$ implicitly as a function of $x\text{.}$ The graph of the equation can be broken into pieces where each piece can be defined by an explicit function of $x\text{.}$ For the circle, we could choose to take the top half as one explicit function of $x\text{,}$ namely $y=\sqrt{16-x^2}$ and the bottom half as the explicit function $y=-\sqrt{16-x^2}\text{.}$ The equation for the circle defines an implicit function of $x\text{.}$

The righthand curve in Figure 2.7.1 is called the folium of Descartes and is just one of many fascinating possibilities for implicitly given curves.

How can we find an equation for $\frac{dy}{dx}$ without an explicit formula for $y$ in terms of $x\text{?}$ To begin answering this question, the following preview activity reminds us of some ways we can compute derivatives of functions in settings where the function’s formula is not known.

We begin our exploration of implicit differentiation with the example of the circle given by $x^2+y^2=16\text{.}$ How can we find a formula for $\frac{dy}{dx}\text{?}$

By viewing $y$ as an implicit function of $x\text{,}$ we think of $y$ as some function whose formula $f(x)$ is unknown, but which we can differentiate. Just as $y$ represents an unknown formula, so too its derivative with respect to $x\text{,}$ $\frac{dy}{dx}\text{,}$ will be (at least temporarily) unknown.

All of these observations about the circle are consistent with the formula $\frac{dy}{dx}=-\frac{x}{y}\text{.}$

Example 2.7.3 shows that it is possible when differentiating implicitly to have multiple terms involving $\frac{dy}{dx}\text{.}$ We use addition and subtraction to collect all terms involving $\frac{dy}{dx}$ on one side of the equation, then factor to get a single term of $\frac{dy}{dx}\text{.}$ Finally, we divide to solve for $\frac{dy}{dx}\text{.}$

The tangent line is horizontal precisely when the numerator is zero and the denominator is nonzero, making the slope of the tangent line zero. If we can solve the equation $p(x,y)=0$ for either $x$ and $y$ in terms of the other, we can substitute that expression into the original equation for the curve. This gives an equation in a single variable, and if we can solve that equation we can find the point(s) on the curve where $p(x,y)=0\text{.}$ At those points, the tangent line is horizontal.

Similarly, the tangent line is vertical whenever $q(x,y)=0$ and $p(x,y)\ne 0\text{,}$ making the slope undefined.