# 2.6: Implicit Differentiation

**At Grade**Created by: CK-12

## Learning Objectives

A student will be able to:

- Find the derivative of variety of functions by using the technique of implicit differentiation.

Consider the equation

We want to obtain the derivative

and then project the derivative on both sides,

There is another way of finding

Using the Product Rule on the left-hand side,

Solving for

But since

which agrees with the previous calculations. This second method is called the **implicit differentiation** method. You may wonder and say that the first method is easier and faster and there is no reason for the second method. That’s probably true, but consider this function:

How would you solve for

**Example 1:**

Find

*Solution:*

Differentiating both sides with respect to

Solving for

Implicit differentiation can be used to calculate the slope of the tangent line as the example below shows.

**Example 2:**

Find the equation of the tangent line that passes through point

*Solution:*

First we need to use implicit differentiation to find

Now, substituting point

So the slope of the tangent line is

Next we need to find the equation of the tangent line. The slope-intercept form is

where

Thus the equation of the tangent line is

**Remark:** we could have used the point-slope form

**Example 3:**

Use implicit differentiation to find

*Solution:*

Solving for

Differentiating both sides implicitly again (and using the quotient rule),

But since

This is the second derivative of \begin{align*}y\end{align*}.

The next step is to find: \begin{align*} \frac{d^2y} {dx^2}\Big|_{(x,~y) = (2,~3)}\end{align*}

\begin{align*}\frac{d^2y} {dx^2} \Bigg |_{(2,~3)} &= \frac{5} {4(3)} - \frac{25(2)^2} {16(3)^3}\\ &= \frac{5} {27}.\end{align*}

Since the first derivative of a function represents the rate of change of the function \begin{align*}y = f(x)\end{align*} with respect to \begin{align*}x\end{align*}, the second derivative represents the rate of change of the rate of change of the function. For example, in kinematics (the study of motion), the speed of an object \begin{align*}(y')\end{align*} signifies the change of position with respect to time but acceleration \begin{align*}(y'')\end{align*} signifies the rate of change of the speed with respect to time.

## Multimedia Links

For more examples of implicit differentiation **(6.0)**, see Math Video Tutorials by James Sousa, Implicit Differentiation (8:10).

For a video presentation of related rates using implicit differentiation **(6.0)**, see Just Math Tutoring, Related Rates Using Implicit Differentiation (9:57).

For a presentation of related rates using cones **(6.0)**, see Just Math Tutoring, Related Rates Using Implicit Differentiation (2:47).

## Review Questions

Find \begin{align*}dy/dx\end{align*} by implicit differentiation.

- \begin{align*}x^2 + y^2 = 500\end{align*}
- \begin{align*}x^2 y + 3 xy - 2 = 1\end{align*}
- \begin{align*} \frac{1} {x} + \frac{1} {y} = \frac{1} {2}\end{align*}
- \begin{align*} \sqrt{x} - \sqrt{y} =\sqrt{3}\end{align*}
- \begin{align*}\sin(25xy^2) = x\end{align*}
- \begin{align*}\tan^3 (x^2 - y^2) = \tan(\pi/4)\end{align*}

In problems #7 and 8, use implicit differentiation to find the slope of the tangent line to the given curve at the specified point.

- \begin{align*}x^2 y - y^2 x = 0\end{align*} at \begin{align*}(1, 1)\end{align*}
- \begin{align*}\sin(xy) = y\end{align*} at \begin{align*}(\frac{\pi}{2}, 1)\end{align*}
- Find \begin{align*}y''\end{align*} by implicit differentiation for \begin{align*}x^3y^3 = 5\end{align*}.
- Use implicit differentiation to show that the tangent line to the curve \begin{align*}y^2 = kx\end{align*} at \begin{align*}(x_0, y_0)\end{align*} is given by \begin{align*} yy_0 = \frac{1} {2}k (x + x_0)\end{align*}, where \begin{align*}k\end{align*} is a constant.

\begin{align*}\frac{d(y^2)}{dx} &= \frac{d(kx)}{dx}\\ 2y\frac{dy}{dx} &= k\\ \frac{dy}{dx} &= \frac{k}{2y}\end{align*}

Second, we substitute \begin{align*}y_0\end{align*} for \begin{align*}y\end{align*}, and that gives us the slope \begin{align*}m\end{align*} of our tangent line at \begin{align*}(x_0, y_0)\end{align*}:

\begin{align*}m = \frac{k}{2y_0}\end{align*}

Third, we set up the equation for our tangent line using point-slope form:

\begin{align*}y - y_0 = \frac{k}{2y_0}(x - x_0)\end{align*}

Fourth, and finally, we manipulate this linear equation to get the term \begin{align*}yy_0\end{align*} isolated on the left hand side:

\begin{align*}y - y_0 &= \frac{k}{2y_0}(x- x_0)\\ y &= \frac{k}{2y_0}(x - x_0) + y_0\\ yy_0 &= \frac{k}{2}(x - x_0)+(y_0)^2\\ yy_0 &= \frac{k}{2}(x - x_0) + kx_0 \ (\text{Using the fact that} \ y^2 = kx)\\ yy_0 &= \frac{k}{2}(x + x_0)\end{align*}

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