Implicit Differentiation

We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specific point. In all these cases we had the explicit equation for the function and differentiated these functions explicitly. Suppose instead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrary curve at a point. In this section, we solve these problems by finding the derivatives of functions that define

y

Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of

y

In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define many different functions implicitly. For example, the functions

and

y = {\begin{matrix} \sqrt{25 - x^{2}} if - 25 \leq x < 0 \\ − \sqrt{25 - x^{2}} if 0 \leq x \leq 25 \end{matrix},

which are illustrated in [link], are just three of the many functions defined implicitly by the equation

x^{2} + y^{2} = 25 .

If we want to find the slope of the line tangent to the graph of

x^{2} + y^{2} = 25

On the other hand, if we want the slope of the tangent line at the point

(3, −4),

However, it is not always easy to solve for a function defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly. The process of finding

\frac{d y}{d x}

using implicit differentiation is described in the following problem-solving strategy.

Using Implicit Differentiation

Assuming that $y$

is defined implicitly by the equation $x^{2} + y^{2} = 25,$

find $\frac{d y}{d x} .$

Follow the steps in the problem-solving strategy.

\begin{matrix} \frac{d}{d x} (x^{2} + y^{2}) & = & \frac{d}{d x} (25) & Step 1. Differentiate both sides of the equation. \\ \frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) & = & 0 & \begin{matrix} Step 1.1. Use the sum rule on the left. \\ On the right \frac{d}{d x} (25) = 0 . \end{matrix} \\ 2 x + 2 y \frac{d y}{d x} & = & 0 & \begin{matrix} Step 1.2. Take the derivatives, so \frac{d}{d x} (x^{2}) = 2 x \\ and \frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x} . \end{matrix} \\ 2 y \frac{d y}{d x} & = & −2 x & \begin{matrix} Step 2. Keep the terms with \frac{d y}{d x} on the left. \\ Move the remaining terms to the right. \end{matrix} \\ \frac{d y}{d x} & = & - \frac{x}{y} & \begin{matrix} Step 4. Divide both sides of the equation by \\ 2 y . (Step 3 does not apply in this case.) \end{matrix} \end{matrix}

Analysis

Note that the resulting expression for $\frac{d y}{d x}$

is in terms of both the independent variable $x$

and the dependent variable $y .$

Although in some cases it may be possible to express $\frac{d y}{d x}$

in terms of $x$

only, it is generally not possible to do so.

Using Implicit Differentiation and the Product Rule

Assuming that $y$

is defined implicitly by the equation $x^{3} sin y + y = 4 x + 3,$

find $\frac{d y}{d x} .$

\begin{matrix} \frac{d}{d x} (x^{3} sin y + y) & = & \frac{d}{d x} (4 x + 3) & Step 1: Differentiate both sides of the equation. \\ \frac{d}{d x} (x^{3} sin y) + \frac{d}{d x} (y) & = & 4 & \begin{matrix} Step 1.1: Apply the sum rule on the left. \\ On the right, \frac{d}{d x} (4 x + 3) = 4 . \end{matrix} \\ (\frac{d}{d x} (x^{3}) \cdot sin y + \frac{d}{d x} (sin y) \cdot x^{3}) + \frac{d y}{d x} & = & 4 & \begin{matrix} Step 1.2: Use the product rule to find \\ \frac{d}{d x} (x^{3} sin y) . Observe that \frac{d}{d x} (y) = \frac{d y}{d x} . \end{matrix} \\ 3 x^{2} sin y + (cos y \frac{d y}{d x}) \cdot x^{3} + \frac{d y}{d x} & = & 4 & \begin{matrix} Step 1.3: We know \frac{d}{d x} (x^{3}) = 3 x^{2} . Use the \\ chain rule to obtain \frac{d}{d x} (sin y) = cos y \frac{d y}{d x} . \end{matrix} \\ x^{3} cos y \frac{d y}{d x} + \frac{d y}{d x} & = & 4 - 3 x^{2} sin y & \begin{matrix} Step 2: Keep all terms containing \frac{d y}{d x} on the \\ left. Move all other terms to the right. \end{matrix} \\ \frac{d y}{d x} (x^{3} cos y + 1) & = & 4 - 3 x^{2} sin y & Step 3: Factor out \frac{d y}{d x} on the left. \\ \frac{d y}{d x} & = & \frac{4 - 3 x^{2} sin y}{x^{3} cos y + 1} & \begin{matrix} Step 4: Solve for \frac{d y}{d x} by dividing both sides of \\ the equation by x^{3} cos y + 1 . \end{matrix} \end{matrix}

Finding Tangent Lines Implicitly

Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations.

Applying Implicit Differentiation

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation $4 x^{2} + 25 y^{2} = 100 .$

The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive x-axis toward $(0, 0) .$

If the rocket fires a missile when it is located at $(3, \frac{8}{3}),$

where will it intersect the x-axis?

To solve this problem, we must determine where the line tangent to the graph of

4 x^{2} + 25 y^{2} = 100

at $(3, \frac{8}{3})$

intersects the x-axis. Begin by finding $\frac{d y}{d x}$

implicitly.

Differentiating, we have

8 x + 50 y \frac{d y}{d x} = 0 .

Solving for $\frac{d y}{d x},$

we have

\frac{d y}{d x} = - \frac{4 x}{25 y} .

The slope of the tangent line is $\frac{d y}{d x} \|_{(3, \frac{8}{3})} = - \frac{9}{50} .$

The equation of the tangent line is $y = - \frac{9}{50} x + \frac{183}{200} .$

To determine where the line intersects the x-axis, solve $0 = - \frac{9}{50} x + \frac{183}{200} .$

The solution is $x = \frac{61}{3} .$

The missile intersects the x-axis at the point $(\frac{61}{3}, 0) .$

Key Concepts

For the following exercises, use implicit differentiation to find $\frac{d y}{d x} .$

x^{2} - y^{2} = 4

6 x^{2} + 3 y^{2} = 12

\frac{d y}{d x} = \frac{−2 x}{y}

x^{2} y = y - 7

3 x^{3} + 9 x y^{2} = 5 x^{3}

\frac{d y}{d x} = \frac{x}{3 y} - \frac{y}{2 x}

x y - cos (x y) = 1

y \sqrt{x + 4} = x y + 8

\frac{d y}{d x} = \frac{y - \frac{y}{2 \sqrt{x + 4}}}{\sqrt{x + 4} - x}

− x y - 2 = \frac{x}{7}

y sin (x y) = y^{2} + 2

\frac{d y}{d x} = \frac{y^{2} cos (x y)}{2 y - sin (x y) - x y cos x y}

{(x y)}^{2} + 3 x = y^{2}

x^{3} y + x y^{3} = −8

\frac{d y}{d x} = \frac{−3 x^{2} y - y^{3}}{x^{3} + 3 x y^{2}}

For the following exercises, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line.

[T] $x^{4} y - x y^{3} = −2, (−1, −1)$

[T] $x^{2} y^{2} + 5 x y = 14, (2, 1)$

The graph has a crescent in each of the four quadrants. There is a straight line marked T(x) with slope −1/2 and y intercept 2.

y = \frac{−1}{2} x + 2

[T] $tan (x y) = y, (\frac{π}{4}, 1)$

[T] $x y^{2} + sin (π y) - 2 x^{2} = 10, (2, −3)$

The graph has two curves, one in the first quadrant and one in the fourth quadrant. They are symmetric about the x axis. The curve in the first quadrant goes from (0.3, 5) to (1.5, 3.5) to (5, 4). There is a straight line marked T(x) with slope 1/(π + 12) and y intercept −(3π + 38)/(π + 12).

y = \frac{1}{π + 12} x - \frac{3 π + 38}{π + 12}

[T] $\frac{x}{y} + 5 x - 7 = - \frac{3}{4} y, (1, 2)$

[T] $x y + sin (x) = 1, (\frac{π}{2}, 0)$

The graph starts in the third quadrant near (−5, 0), remains near 0 until x = −4, at which point it decreases until it reaches near (0, −5). There is an asymptote at x = 0. The graph begins again near (0, 5) decreases to (1, 0) and then increases a little bit before decreasing to be near (5, 0). There is a straight line marked T(x) that coincides with y = 0.

y = 0

[T] The graph of a folium of Descartes with equation $2 x^{3} + 2 y^{3} - 9 x y = 0$

is given in the following graph.

A folium is graphed which has equation 2x3 + 2y3 – 9xy = 0. It crosses over itself at (0, 0).

Find the equation of the tangent line at the point $(2, 1) .$
Graph the tangent line along with the folium.
Find the equation of the normal line to the tangent line in a. at the point $(2, 1) .$

For the equation $x^{2} + 2 x y - 3 y^{2} = 0,$

Find the equation of the normal to the tangent line at the point $(1, 1) .$
At what other point does the normal line in a. intersect the graph of the equation?

a. $y = − x + 2$

b. $(3, −1)$

Find all points on the graph of $y^{3} - 27 y = x^{2} - 90$

at which the tangent line is vertical.

For the equation $x^{2} + x y + y^{2} = 7,$

Find the $x$
-intercept(s).
Find the slope of the tangent line(s) at the x-intercept(s).
What does the value(s) in b. indicate about the tangent line(s)?

a. $(\pm \sqrt{7}, 0)$

b. $−2$

c. They are parallel since the slope is the same at both intercepts.

Find the equation of the tangent line to the graph of the equation ${sin}^{−1} x + {sin}^{−1} y = \frac{π}{6}$

at the point $(0, \frac{1}{2}) .$

Find the equation of the tangent line to the graph of the equation ${tan}^{−1} (x + y) = x^{2} + \frac{π}{4}$

at the point $(0, 1) .$

y = − x + 1

Find $y^{'}$

and $y ″$

for $x^{2} + 6 x y - 2 y^{2} = 3 .$

[T] The number of cell phones produced when $x$

dollars is spent on labor and $y$

dollars is spent on capital invested by a manufacturer can be modeled by the equation $60 x^{3 / 4} y^{1 / 4} = 3240 .$

Find $\frac{d y}{d x}$
and evaluate at the point
$(81, 16) .$
Interpret the result of a.

a. $−0.5926$

b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.

[T] The number of cars produced when $x$

dollars is spent on labor and $y$

dollars is spent on capital invested by a manufacturer can be modeled by the equation $30 x^{1 / 3} y^{2 / 3} = 360 .$

(Both $x$

and $y$

are measured in thousands of dollars.)

Find $\frac{d y}{d x}$
and evaluate at the point
$(27, 8) .$
Interpret the result of a.

The volume of a right circular cone of radius $x$

and height $y$

is given by $V = \frac{1}{3} π x^{2} y .$

Suppose that the volume of the cone is $85 π {cm}^{3} .$

Find $\frac{d y}{d x}$

when $x = 4$

and $y = 16 .$

−8

For the following exercises, consider a closed rectangular box with a square base with side $x$

and height $y .$

Find an equation for the surface area of the rectangular box, $S (x, y) .$

If the surface area of the rectangular box is 78 square feet, find $\frac{d y}{d x}$

when $x = 3$

feet and $y = 5$

feet.

−2.67

For the following exercises, use implicit differentiation to determine $y^{'} .$

Does the answer agree with the formulas we have previously determined?

x = sin y

x = cos y

y^{'} = - \frac{1}{\sqrt{1 - x^{2}}}

x = tan y

Implicit Differentiation