# How do you differentiate f(x)=sqrt(e^cot(x))  using the chain rule?

##### 1 Answer
Jan 5, 2016

f'(x)==-$\frac{\sqrt{{e}^{\cot} \left(x\right)} . {\csc}^{2} \left(x\right)}{2}$

#### Explanation:

$f \left(x\right) = \sqrt{{e}^{\cot} \left(x\right)}$
To find the derivative of f(x), we need to use chain rule .

$\textcolor{red}{\text{chain rule :f(g(x))'=f'(g(x)).g'(x)}}$
Let $u \left(x\right) = \cot \left(x\right) \implies u ' \left(x\right) = - {\csc}^{2} \left(x\right)$
and $g \left(x\right) = {e}^{x} \implies g ' \left(x\right) = {e}^{x} . g ' \left(u \left(x\right)\right) = {e}^{\cot} \left(x\right)$
f(x)=sqrt(x)=>f'(x)=1/(2sqrt(x))=>f'(g(u(x)))=1/(2sqrt(e^cot(x))

d/dx(f(g(u(x)))=f'(g(u(x))).g'(u(x)).u'(x)
=$\frac{1}{\sqrt{{e}^{\cot} \left(x\right)}} {e}^{\cot} \left(x\right) . - {\cos}^{2} \left(x\right)$
=$\frac{- {e}^{\cot} \left(x\right) {\csc}^{2} x}{\sqrt{{e}^{\cot} \left(x\right)}}$
$\textcolor{b l u e}{\text{cancel the e^cot(x) with sqrt(e^cot(x)) in the denominator}}$
=-$\frac{\sqrt{{e}^{\cot} \left(x\right)} . {\csc}^{2} \left(x\right)}{2}$