How do you differentiate f(x)=sqrt(cote^(4x) using the chain rule.?

Mar 5, 2018

$f ' \left(x\right) = \frac{- 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right) {\left(\cot \left({e}^{4 x}\right)\right)}^{- \frac{1}{2}}}{2}$
color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x))

Explanation:

$f \left(x\right) = \sqrt{\cot \left({e}^{4 x}\right)}$
$\textcolor{w h i t e}{f \left(x\right)} = \sqrt{g \left(x\right)}$
$f ' \left(x\right) = \frac{1}{2} \cdot {\left(g \left(x\right)\right)}^{- \frac{1}{2}} \cdot g ' \left(x\right)$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{g ' \left(x\right) {\left(g \left(x\right)\right)}^{- \frac{1}{2}}}{2}$

$g \left(x\right) = \cot \left({e}^{4 x}\right)$
$\textcolor{w h i t e}{g \left(x\right)} = \cot \left(h \left(x\right)\right)$
$g ' \left(x\right) = - h ' \left(x\right) {\csc}^{2} \left(h \left(x\right)\right)$

$h \left(x\right) = {e}^{4 x}$
$\textcolor{w h i t e}{h \left(x\right)} = {e}^{j \left(x\right)}$
$h ' \left(x\right) = j ' \left(x\right) {e}^{j \left(x\right)}$

$j \left(x\right) = 4 x$
$j ' \left(x\right) = 4$

$h ' \left(x\right) = 4 {e}^{4 x}$

$g ' \left(x\right) = - 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right)$

$f ' \left(x\right) = \frac{- 4 {e}^{4 x} {\csc}^{2} \left({e}^{4 x}\right) {\left(\cot \left({e}^{4 x}\right)\right)}^{- \frac{1}{2}}}{2}$
color(white)(f'(x))=-(2e^(4x)csc^2(e^(4x)))/sqrt(cot(e^(4x))