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Implicit Differentiation

Implicit Differentiation Online Tutoring / Homework Help
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 dy/dx in case of implicit functions as well.

Example 1:

Find dy/dx when x2 + y3 + xy = 16

Solution:

Given: x2 + y3 + xy = 16

Differentiating both sides with respect to x we get,

2x + 3y2 dy/dx+ x dy/dx + y(1) = 0

Note:

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

(3y2 + x) dy/dx = -(2x + y)

dy/dx=-(2x+y)/(3y2+x)

Example 2:

Find dy/dx for ex + ey = ex + y

Solution:

ex + ey = ex + y

Differentiate both sides with respect to x

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

ex + ey = e(x + y)(1 + dy/dx )

Now factor out the dy/dx term

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

Finally bring dy/dx to one side of the equation.

dy/dx


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 dy/dx wherever we need to differentiate y term.
  • Factor out the dy/dx terms.
  • Place the dy/dx term on one side of the equation, bringing the remaining terms to the other side.


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