# Trigonometric Differentiation Problem.

• February 8th 2013, 06:43 AM
KhanDisciple
Trigonometric Differentiation Problem.
Hi, for the life of me I cannot figure this one out.

Find the derivative of $sin^4 \sqrt{u}$.

This is what I tried to do:
$sin^4 \sqrt{u} = (sin\sqrt{u})^4$
$=4(sin\sqrt{u})^3 [cos\sqrt{u}({\textstyle \frac{1}{2}}u^\frac{-1}{2})]$

The answer in the book is $2\sqrt{u}sin^3 \sqrt{u} cos\sqrt{u}$.
How the **** did they get that? Thanks.
• February 8th 2013, 07:01 AM
Plato
Re: Trigonometric Differentiation Problem.
Quote:

Originally Posted by KhanDisciple
Hi, for the life of me I cannot figure this one out.

Find the derivative of $sin^4 \sqrt{u}$.

This is what I tried to do:
$sin^4 \sqrt{u} = (sin\sqrt{u})^4$
$=4(sin\sqrt{u})^3 [cos\sqrt{u}({ \frac{1}{2}}u^\frac{-1}{2})]$

The answer in the book is $2\sqrt{u}sin^3 \sqrt{u} cos\sqrt{u}$.
How the **** did they get that? Thanks.

Note that $4\sin^3 (\sqrt{u}) \left[cos(\sqrt{u})\left(\frac{1}{2\sqrt{u}}\right) \right]$ is the correct answer.

Also, because $\sin ~\&~\cos$ are functions please use function notation, like $\sin(x) ~\&~\cos(x)$
• February 8th 2013, 07:36 AM
KhanDisciple
Re: Trigonometric Differentiation Problem.
Thank you Plato. The answers appeared completely different to me, and in the future I will try to use the correct notation.