# How do you differentiate f(x)=sqrtcos(e^(4x)) using the chain rule.?

May 25, 2016

$f ' \left(x\right) = \frac{- 2 \sin \left({e}^{4 x}\right) {e}^{4 x}}{\sqrt{\cos \left({e}^{4 x}\right)}}$

#### Explanation:

Let

• $p = 4 x$
• $q = {e}^{p}$
• $r = \cos \left(q\right)$
• $s = \sqrt{r}$

Therefore, to differentiate $f \left(x\right)$, we need to find $f ' \left(x\right)$, which is the same as frac{"d"s}{"d"x}.

Applying the chain rule,

frac{"d"s}{"d"x} = frac{"d"s}{"d"r} frac{"d"r}{"d"q} frac{"d"q}{"d"p} frac{"d"p}{"d"x}

$= \frac{1}{2 \sqrt{r}} \cdot \left(- \sin \left(q\right)\right) \cdot {e}^{p} \cdot \left(4\right)$

$= \frac{- 2 \sin \left({e}^{4 x}\right) {e}^{4 x}}{\sqrt{\cos \left({e}^{4 x}\right)}}$