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

##### 1 Answer

$f ' \left(x\right) = - \setminus \frac{{e}^{\frac{1}{x}} \cos e {c}^{2} \left({e}^{\frac{1}{x}}\right)}{{x}^{2}}$

#### Explanation:

Given function:

$f \left(x\right) = \setminus \cot \left({e}^{\frac{1}{x}}\right)$

differentiating above function w.r.t $x$ using chain rule as follows

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

$f ' \left(x\right) = - \cos e {c}^{2} \left({e}^{\frac{1}{x}}\right) \setminus \frac{d}{\mathrm{dx}} \left({e}^{\frac{1}{x}}\right)$

$= - \cos e {c}^{2} \left({e}^{\frac{1}{x}}\right) \setminus \cdot {e}^{\frac{1}{x}} \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

$= - \cos e {c}^{2} \left({e}^{\frac{1}{x}}\right) \setminus \cdot {e}^{\frac{1}{x}} \left(- \frac{1}{x} ^ 2\right)$

$= - \setminus \frac{{e}^{\frac{1}{x}} \cos e {c}^{2} \left({e}^{\frac{1}{x}}\right)}{{x}^{2}}$