# How do you differentiate f(x)=sqrt(e^(cot(1/x) using the chain rule.?

Sep 1, 2017

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

#### Explanation:

Firstly, differentiate the outermost function, which is $\sqrt{\ldots}$

$\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$

Hence, $\frac{d}{\mathrm{dx}} \left(\sqrt{{e}^{\cot \left(\frac{1}{x}\right)}}\right) = \frac{1}{2 \sqrt{{e}^{\cot \left(\frac{1}{x}\right)}}} \cdot \frac{d}{\mathrm{dx}} \left({e}^{\cot \left(\frac{1}{x}\right)}\right)$

Now apply the chain rule again to find $\frac{d}{\mathrm{dx}} \left({e}^{\cot \left(\frac{1}{x}\right)}\right)$

$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

$\frac{d}{\mathrm{dx}} \left({e}^{\cot \left(\frac{1}{x}\right)}\right) = {e}^{\cot \left(\frac{1}{x}\right)} \cdot \frac{d}{\mathrm{dx}} \left(\cot \left(\frac{1}{x}\right)\right)$

Now apply the chain rule one more time

$\frac{d}{\mathrm{dx}} \left(\cot \left(x\right)\right) = - {\csc}^{2} \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(\cot \left(\frac{1}{x}\right)\right) = - {\csc}^{2} \left(\frac{1}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

Finally, find the derivative of the innermost function

$\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right) = - \frac{1}{x} ^ 2$

Now just put everything together

$\frac{d}{\mathrm{dx}} \left(\sqrt{{e}^{\cot \left(\frac{1}{x}\right)}}\right) = \frac{1}{2 \sqrt{{e}^{\cot \left(\frac{1}{x}\right)}}} \cdot {e}^{\cot \left(\frac{1}{x}\right)} \cdot \left(- {\csc}^{2} \left(\frac{1}{x}\right)\right) \cdot \left(- \frac{1}{x} ^ 2\right)$

After simplifying,

$\frac{d}{\mathrm{dx}} \left(\sqrt{{e}^{\cot \left(\frac{1}{x}\right)}}\right) = \frac{\sqrt{{e}^{\cot \left(\frac{1}{x}\right)}} {\csc}^{2} \left(\frac{1}{x}\right)}{2 {x}^{2}}$