 Power Rule in Differentiation (Part 3) How to solve the power rule in differentiation problems (when the exponent is a rational number): formula, proof, examples, and their solutions.

Formula Previously, you've learned that
[xn]' = nxn - 1
when n is an integer.

Power rule in differentiation (Part 1)

Power rule in differentiation (Part 2)

This is also true
when n is a rational number: a fraction.

Proof Set n = a/b.

Then y = xn can be changed into y = xa/b.

Raise both sides to the power of b.

Then yb = xa.

Differentiate both sides.

[yb]' = byb - 1y'

[xa]' = axa - 1

Implicit differentiation

To make [y' = ...] form,
divide both sides by byb - 1.

Put xa/b into the denominator's y.

Expand the denominator's exponent.

Cancel a and -a. (dark gray terms)

Change a/b into n.

Then [xn]' = nxn - 1
when n is a rational number.

Example 1 The square root is the power of 1/2.

So y = x1/2.

Rational exponents

Use the power rule to find y'.

y' = (1/2)x1/2 - 1

So y' = 1/2√x.

The derivative of √x is frequently used.
So it's good to remember that [√x]' = 1/2√x.

If your teacher wants you
to rationalize the denominator,

rationalize the denominator
by multiplying √x/√x.

Example 2 Change the radical into exponent form.

Then y = x-3/4.

Rational exponents

Use the power rule to find y'.

y' = (-3/4)x-3/4 - 1

So y' = -3/4x4x.

If your teacher wants you
to rationalize the denominator,

rationalize the denominator
by multiplying 4x/4x.

Example 3 Change the radical into exponent form.

Then y = (2x - 1)4/3.

Rational exponents

Use the power rule and chain rule to find y'.