# Implicit Differentiation

Introduction

We have worked with functions where y can be expressed explicitly in terms of x.
For example, lets take a look at the following functions

1. x + y = 2 which can be written as y = 2  x.
2. y = x2
3. y = tan x + ln x

All the above functions are explicit functions.
But sometimes it is not as simple to write y in terms of x.

For example,

1. x2 + y3 + xy = 2
2. tan(x + y)  tan(x  y) = 1
As the above functions cannot be explicitly written in terms of x we call them implicit functions.

Now, it is possible to find in case of implicit functions as well.

Example 1:

Find when x2 + y3 + xy = 16

Solution:

Given: x2 + y3 + xy = 16

Differentiating both sides with respect to x we get,

2x + 3y2 + x + y(1) = 0

Note:

1. We differentiated y3 by using the power rule and attaching
2. Differentiate xy by using the product rule.
Next, bring to one side of the equation as follows:

(3y2 + x) = -(2x + y)

Example 2:

Find for ex + ey = ex + y

Solution:

ex + ey = ex + y

Differentiate both sides with respect to x

ex + ey = e(x + y) (x + y) . Applying chain rule on the right side

ex + ey = e(x + y)(1 + )

Now factor out the term

( ey - e(x + y)) = e(x + y) - ex

Finally bring to one side of the equation.

Summary:

We differentiate the implicit functions by applying the following steps:

• Differentiate both sides of the equation with respect to x using the regular differentiation rules.
• Attach wherever we need to differentiate y term.
• Factor out the terms.
• Place the term on one side of the equation, bringing the remaining terms to the other side.

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