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

Jan 9, 2016

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

Explanation:

First Issue: The $e$ power. According to the chain rule, $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$. Thus,

$f ' \left(x\right) = {e}^{\cot} \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \left(\cot \sqrt{x}\right)$

Second Issue: the cotangent function. Again through the chain rule, $\frac{d}{\mathrm{dx}} \left(\cot u\right) = - {\csc}^{2} u \cdot u '$. Thus,

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

Now, to differentiate $\sqrt{x}$, treat it as ${x}^{\frac{1}{2}}$. Thus, its derivative is $\frac{1}{2} {x}^{- \frac{1}{2}}$ or $\frac{1}{2 \sqrt{x}}$.

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