Calculus:Power Rule

Power Rule

$(x^{n})'=n*x^{n-1}$

Hint

n can be any number that is a kind of constant $1 \over 2$ is constant.

also when you have square root

${\sqrt {x}}$

you can look it as

$x^{1 \over 2}$

Examples

Example 1

EX1:$f'(x^{6})$

Using the Power rule $f'(x^{n})=n*x^{n-1}$

$f'(x^{6})=6*x^{6-1}$

$f'(x^{6})=6x^{5}$

Example 2

EX2:$f'(x^{7})$

Using the Power rule $f'(x^{n})=n*x^{n-1}$

$f'(x^{7})=7*x^{7-1}$

$f'(x^{7})=7x^{6}$

Example 3

EX3:$f'(x^{7})$

Using the Power rule $f'(x^{n})=n*x^{n-1}$

$f'(x^{7})=7*x^{7-1}$

$f'(x^{7})=7x^{6}$

Example 4

EX4:$f'({x^{7} \over 7})$

Using the Power rule $f'(x^{n})=n*x^{n-1}$

$f'({x^{7} \over 7})={1 \over 7}7*x^{7-1}$

$f'({x^{7} \over 7})=x^{6}$

Example 5

EX4:$f'({\sqrt[{4}]{x}})$

first turn it in to exponent $f'({\sqrt[{4}]{x}})=f'({x^{1 \over 4}})$

use the power rule

$f'({\sqrt[{4}]{x}})=({x^{{1 \over 4}-{1}}})$

$f'({\sqrt[{4}]{x}})=({x^{{1 \over 4}-{1}}})$

$f'({\sqrt[{4}]{x}})=({x^{-3 \over 4}})$