# How do you find f' if f(x)=cos^2(3sqrt x )?

Mar 26, 2018

$f ' \left(x\right) = \frac{- 3 \cos \left(3 \sqrt{x}\right) \sin \left(3 \sqrt{x}\right)}{\sqrt{x}}$

#### Explanation:

One can use the chain rule break the function into two parts $u = 3 \sqrt{x} \mathmr{and} 3 {x}^{0.5}$ and $f \left(u\right) = {\left(\cos \left(u\right)\right)}^{2}$

Then derive each part ${x}^{n}$ becomes $n {x}^{n - 1}$ and $\cos \left(x\right)$ becomes $- \sin \left(x\right)$ so

$u ' = 1.5 {x}^{-} 0.5$
and $f ' \left(u\right) = 2 \cos \left(u\right) \cdot - \sin \left(u\right)$ as we have a function inside a function ${\left(\right)}^{2}$ and $\cos \left(\right)$ inside it so we have to derive it the outside function and multiply it by the inside function

Then all we have to do now is take replace the $u$'s with $3 \sqrt{x}$ and multiply the very inside derivative $u '$ by $f ' \left(u\right)$

$f ' \left(x\right) = \frac{- 3 \cos \left(3 \sqrt{x}\right) \sin \left(3 \sqrt{x}\right)}{\sqrt{x}}$