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

Feb 17, 2016

$\frac{1}{e} ^ \left(x\right) {\csc}^{2} \left(\frac{1}{e} ^ x\right)$

Explanation:

To do this we need the chain rule.

First we find the derivative of the function inside the cot function. Remember $\frac{1}{e} ^ x = {e}^{-} x$

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

Also the derivative of $\cot \left(x\right)$:
$\frac{d}{\mathrm{dx}} \cot \left(x\right) = - {\csc}^{2} \left(x\right)$

The chain rule states that: $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = g ' \left(x\right) f ' \left(g \left(x\right)\right)$
So we take the derivative of the inside and multiply it by the derivative of the outside. We found the derivatives, we just need to multiply to get:

$\frac{1}{e} ^ \left(x\right) {\csc}^{2} \left(\frac{1}{e} ^ x\right)$