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

Feb 9, 2016

#### Answer:

[2e^(4x)cos(e(^4x))] / [sqrt sin(e^(4x)]

#### Explanation:

Applying chain rule
df(u)/dx= df/du .du/dx
let sin e^(4x) =u
d/du √u . d/dx (sin(e^(4x)))
we have,
d/du √u=1/(2√u)
and
d/dx (sin(e^(4x)))

Applying chain rule,
df(u)/dx= df/du .du/dx
let ${e}^{4 x} = u$
$\setminus \frac{d}{\mathrm{du}}$ (sin(u) $\setminus \frac{d}{\mathrm{dx}}$

$\setminus \frac{d}{\mathrm{du}}$ sin(u) = cos(u)
$\setminus \frac{d}{\mathrm{dx}} \left({e}^{4 x} \setminus\right)$
Applying chain rule,
df(u)/dx= df/du .du/dx
let 4x=u
solving it we get,
${e}^{4 x}$4
now, cos(u) ${e}^{4 x} 4$
u= ${e}^{4 x}$
= cos(${e}^{4 x}$) ${e}^{4 x} 4$

Finally,
$\setminus \frac{1}{2 \setminus \sqrt{u}}$ cos(${e}^{4 x}$) ${e}^{4 x} 4$
Substitute u= sin( $\left({e}^{4 x}\right)$
=$\setminus \frac{1}{2 \setminus \sqrt{\setminus \sin \left({e}^{4 x}\right)}}$ cos( $\left({e}^{4 x}\right) {e}^{4 x} 4$
simplifying we get,
$\setminus \frac{2 {e}^{4 x} \setminus \cos \left({e}^{4 x}\right)}{\setminus \sqrt{\setminus \sin \left({e}^{4 x}\right)}}$