# How do you differentiate f(x) = sqrt(sin(4x)  using the chain rule?

Nov 3, 2016

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

#### Explanation:

$f \left(x\right)$ is a composite function of three functions

$\textcolor{b l u e}{u \left(x\right) = \sqrt{x} \mathmr{and} v \left(x\right) = \sin x \mathmr{and} w \left(x\right) = 4 x}$

$f \left(x\right) = u \left(v \left(w \left(x\right)\right)\right) = u \circ \left(v \circ w\right) \left(x\right)$

differentiation of $f \left(x\right)$ or any composite function is determined by applying the chain rule that follows;

f'(x)=(u@(v@w)(x))'=(u(v@w(x))))'=u'(v@w(x)))xx(v@w(x))'=u'(v@w(x)))xx(v'(w(x))xxw'(x)

So,
color(red)(f'(x)=u'(v@w(x))xx(v'(w(x))xxw'(x))

$u ' \left(x\right) = \frac{1}{2 \sqrt{x}}$
$u ' \left(v \circ w \left(x\right)\right) = \frac{1}{2 \sqrt{v \left(w \left(x\right)\right)}} = \frac{1}{2 \sqrt{\sin 4 x}}$
Then,
$\textcolor{red}{u ' \left(v \circ w \left(x\right)\right) = \frac{1}{2 \sqrt{\sin 4 x}}}$

$v ' \left(x\right) = \cos x$
$v ' \left(w \left(x\right)\right) = \cos \left(w \left(x\right)\right)$
Then,
$\textcolor{red}{v ' \left(w \left(x\right)\right) = \cos 4 x}$

color(red)(w'(x)=4

color(red)(f'(x)=u'(v@w(x))xx(v'(w(x))xxw'(x))
$f ' \left(x\right) = \frac{1}{2 \sqrt{\sin 4 x}} \times \cos 4 x \times 4$
$f ' \left(x\right) = \frac{4 \cos 4 x}{2 \sqrt{\sin 4 x}}$

Therefore,
$f ' \left(x\right) = \frac{2 \cos 4 x}{\sqrt{\sin 4 x}}$