Trigonometric Differentiation Problem.

Hi, for the life of me I cannot figure this one out.

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

This is what I tried to do:

$\displaystyle sin^4 \sqrt{u} = (sin\sqrt{u})^4$

$\displaystyle =4(sin\sqrt{u})^3 [cos\sqrt{u}(${\textstyle \frac{1}{2}}u^\frac{-1}{2})]$

The answer in the book is $\displaystyle 2\sqrt{u}sin^3 \sqrt{u} cos\sqrt{u}$.

How the **** did they get that? Thanks.

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 $\displaystyle sin^4 \sqrt{u}$.

This is what I tried to do:

$\displaystyle sin^4 \sqrt{u} = (sin\sqrt{u})^4$

$\displaystyle =4(sin\sqrt{u})^3 [cos\sqrt{u}({ \frac{1}{2}}u^\frac{-1}{2})]$

The answer in the book is $\displaystyle 2\sqrt{u}sin^3 \sqrt{u} cos\sqrt{u}$.

How the **** did they get that? Thanks.

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

Also, because $\displaystyle \sin ~\&~\cos$ are **functions** please use function notation, like $\displaystyle \sin(x) ~\&~\cos(x)$

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.