# chain rule again

• Feb 16th 2008, 05:05 PM
b00yeah05
chain rule again
hey there..just studying for the exam,,im pretty sure im correct but i just wanted to make sure....

differentiate --- $(sqrt(x^3+5))$

by using the chain rule i get my answer to be....

$(1/2(x^3+5)^(-1/2)) . 3x^2$

$3x^2 / 2sqrt(x^3+5)$

Are they the same??? my answer and the proper answer?
• Feb 16th 2008, 05:08 PM
topsquark
Quote:

Originally Posted by b00yeah05
hey there..just studying for the exam,,im pretty sure im correct but i just wanted to make sure....

differentiate --- $(sqrt(x^3+5))$

by using the chain rule i get my answer to be....

$(1/2(x^3+5)^(-1/2)) . 3x^2$

$3x^2 / 2sqrt(x^3+5)$

Are they the same??? my answer and the proper answer?

They are the same.
$\frac{1}{\sqrt{x}} = x^{-1/2}$

-Dan
• Feb 16th 2008, 05:16 PM
Soroban
Hello, b00yeah05!

You need to polish up your algebra . . .

Quote:

Differentiate: . $f(x)\:=\:\sqrt{x^3+5}$

By using the chain rule i get my answer to be: . $f'(x) \:=\:\frac{1}{2}(x^3+5)^{-\frac{1}{2}}\cdot 3x^2$

The answer says: . $\frac{3x^2}{2\sqrt{x^3+5}}$

Are they the same? . . . . . You really can't tell?

Your answer is: . $\frac{1}{2}(x^3+5)^{-\frac{1}{2}}\cdot3x^2 \;=\;\frac{1}{2}\cdot\frac{1}{(x^3+5)^{\frac{1}{2} }}\cdot 3x^2 \;= \;\frac{3x^2}{2(x^3+5)^{\frac{1}{2}}} \;=\;\frac{3x^2}{2\sqrt{x^3+5}}$

• Feb 16th 2008, 05:17 PM
b00yeah05
no i really cant, i really get confused with powers...its not my thing...