2.7: Derivatives of Functions Given Implicitly

Implicit Differentiation

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.

So we view $y$ as an unknown differentiable function of $x$ and differentiate both sides of the equation with respect to $x\text{.}$

$$\frac{d}{dx} \left[ x^2 + y^2 \right] = \frac{d}{dx} \left[ 16 \right]\text{.}$$

On the right, the derivative of the constant $16$ is $0\text{,}$ and on the left we can apply the sum rule, so it follows that

Note carefully the different roles being played by $x$ and $y\text{.}$ Because $x$ is the independent variable, $\frac{d}{dx} \left[x^2\right] = 2x\text{.}$ But $y$ is the dependent variable and $y$ is an implicit function of $x\text{.}$ Recall Preview Activity $2.7.1$ , where we computed $\frac{d}{dx}[f(x)^2]\text{.}$ Computing $\frac{d}{dx}[y^2]$ is the same, and requires the chain rule, by which we find that $\frac{d}{dx}[y^2] = 2y^1 \frac{dy}{dx}\text{.}$ We now have that

We solve this equation for $\frac{dy}{dx}$ by subtracting $2x$ from both sides and dividing by $2y\text{.}$

There are several important things to observe about the result that $\frac{dy}{dx} = -\frac{x}{y}\text{.}$ First, this expression for the derivative involves both $x$ and $y\text{.}$ This makes sense because there are two corresponding points on the circle for each value of $x$ between $-4$ and $4\text{,}$ and the slope of the tangent line is different at each of these points.

Example $2.7.1$

For the curve given implicitly by $x^3 + y^2 - 2xy = 2\text{,}$ shown in Figure $2.7.3$ , find the slope of the tangent line at $(-1,1)\text{.}$

Answer

We begin by differentiating the curve's equation implicitly. Taking the derivative of each side with respect to $x\text{,}$

$$\frac{d}{dx}\left[ x^3 + y^2 - 2xy \right] = \frac{d}{dx} \left[ 2 \right]\text{,}$$

by the sum rule and the fact that the derivative of a constant is zero, we have

$$\frac{d}{dx}[x^3] + \frac{d}{dx}[y^2] - \frac{d}{dx}[2xy] = 0\text{.}$$

For the three derivatives we now must execute, the first uses the simple power rule, the second requires the chain rule (since $y$ is an implicit function of $x$), and the third necessitates the product rule (again since $y$ is a function of $x$). Applying these rules, we now find that

$$3x^2 + 2y\frac{dy}{dx} - [2x \frac{dy}{dx} + 2y] = 0\text{.}$$

We want to solve this equation for $\frac{dy}{dx}\text{.}$ To do so, we first collect all of the terms involving $\frac{dy}{dx}$ on one side of the equation.

$$2y\frac{dy}{dx} - 2x \frac{dy}{dx}= 2y - 3x^2\text{.}$$

Then we factor the left side to isolate $\frac{dy}{dx}\text{.}$

$$\frac{dy}{dx}(2y - 2x) = 2y - 3x^2\text{.}$$

Finally, we divide both sides by $(2y - 2x)$ and conclude that

$$\frac{dy}{dx} = \frac{2y-3x^2}{2y-2x}\text{.}$$

Note that the expression for $\frac{dy}{dx}$ depends on both $x$ and $y\text{.}$ To find the slope of the tangent line at $(-1,1)\text{,}$ we substitute the coordinates into the formula for $\frac{dy}{dx}\text{,}$ using the notation

$$\left. \frac{dy}{dx} \right|_{(-1,1)} = \frac{2(1)-3(-1)^2}{2(1)-2(-1)} = -\frac14\text{.}$$

This value matches our visual estimate of the slope of the tangent line shown in Figure $2.7.3$ .

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 $\frac{dy}{dx}\text{.}$

We use the notation

to denote the evaluation of $\frac{dy}{dx}$ at the point $(a,b)\text{.}$ This is analogous to writing $f'(a)$ when $f'$ depends on a single variable.

There is a big difference between writing $\frac{d}{dx}$ and $\frac{dy}{dx}\text{.}$ For example,

gives an instruction to take the derivative with respect to $x$ of the quantity $x^2 + y^2\text{,}$ presumably where $y$ is a function of $x\text{.}$ On the other hand,

means the product of the derivative of $y$ with respect to $x$ with the quantity $x^2 + y^2\text{.}$ Understanding this notational subtlety is essential.

Activity $2.7.1$

Consider the curve defined by the equation $x = y^5 - 5y^3 + 4y\text{,}$ whose graph is pictured in Figure $2.7.4$ .

1. Explain why it is not possible to express $y$ as an explicit function of $x\text{.}$
2. Use implicit differentiation to find a formula for $dy/dx\text{.}$
3. Use your result from part (b) to find an equation of the line tangent to the graph of $x = y^5 - 5y^3 + 4y$ at the point $(0, 1)\text{.}$
4. Use your result from part (b) to determine all of the points at which the graph of $x = y^5 - 5y^3 + 4y$ has a vertical tangent line.

It is natural to ask where the tangent line to a curve is vertical or horizontal. The slope of a horizontal tangent line must be zero, while the slope of a vertical tangent line is undefined. Often the formula for $\frac{dy}{dx}$ is expressed as a quotient of functions of $x$ and $y\text{,}$ say

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.

Activity $2.7.2$

Consider the curve defined by the equation $y(y^2-1)(y-2) = x(x-1)(x-2)\text{,}$ whose graph is pictured in Figure $2.7.5$ . Through implicit differentiation, it can be shown that

$$\frac{dy}{dx} = \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{(y^2-1)(y-2) + 2y^2(y-2) + y(y^2-1)}\text{.}$$

Use this fact to answer each of the following questions.

1. Determine all points $(x,y)$ at which the tangent line to the curve is horizontal. (Use technology appropriately to find the needed zeros of the relevant polynomial function.)
2. Determine all points $(x,y)$ at which the tangent line is vertical. (Use technology appropriately to find the needed zeros of the relevant polynomial function.)
3. Find the equation of the tangent line to the curve at one of the points where $x = 1\text{.}$

Activity $2.7.3$

For each of the following curves, use implicit differentiation to find $dy/dx$ and determine the equation of the tangent line at the given point.

1. $x^3 - y^3 = 6xy\text{,}$ $(-3,3)$
2. $\sin(y) + y = x^3 + x\text{,}$ $(0,0)$
3. $3x e^{-xy} = y^2\text{,}$ $(0.619061,1)$

Summary

• In an equation involving $x$ and $y$ where portions of the graph can be defined by explicit functions of $x\text{,}$ we say that $y$ is an implicit function of $x\text{.}$ A good example of such a curve is the unit circle.
• We use implicit differentiation to differentiate an implicitly defined function. We differentiate both sides of the equation with respect to $x\text{,}$ treating $y$ as a function of $x$ by applying the chain rule. If possible, we subsequently solve for $\frac{dy}{dx}$ using algebra.
• While $\frac{dy}{dx}$ may now involve both the variables $x$ and $y\text{,}$ $\frac{dy}{dx}$ still gives the slope of the tangent line to the curve. It may be used to decide where the tangent line is horizontal ($\frac{dy}{dx} = 0$) or vertical ($\frac{dy}{dx}$ is undefined), or to find the equation of the tangent line at a particular point on the curve.