# How do you find the derivative of cos((1-e^(2x))/(1+e^(2x)))?

Apr 1, 2018

$f ' \left(x\right) = \frac{4 {e}^{2 x}}{1 + {e}^{2 x}} ^ 2 \sin \left(\frac{1 - {e}^{2 x}}{1 + {e}^{2 x}}\right)$

#### Explanation:

We are dealing with the quotient rule inside the chain rule

Chain rule for cosine

$\cos \left(s\right) \Rightarrow s ' \cdot - \sin \left(s\right)$

Now we have to do the quotient rule

$s = \frac{1 - {e}^{2 x}}{1 + {e}^{2 x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \frac{u}{v} = \frac{u ' v - v ' u}{v} ^ 2$

Rule for deriving e

Rule: ${e}^{u} \Rightarrow u ' {e}^{u}$

Derive both the top and bottom functions

$1 - {e}^{2 x} \Rightarrow 0 - 2 {e}^{2 x}$

$1 + {e}^{2 x} \Rightarrow 0 + 2 {e}^{2 x}$

Put it into the quotient rule

$s ' = \frac{u ' v - v ' u}{v} ^ 2 = \frac{- 2 {e}^{2 x} \left(1 + {e}^{2 x}\right) - 2 {e}^{2 x} \left(1 - {e}^{2 x}\right)}{1 + {e}^{2 x}} ^ 2$

Simply
$s ' = \frac{- 2 {e}^{2 x} \left(\left(1 + {e}^{2 x}\right) + \left(1 - {e}^{2 x}\right)\right)}{1 + {e}^{2 x}} ^ 2$

$s ' = \frac{- 2 {e}^{2 x} \left(2\right)}{1 + {e}^{2 x}} ^ 2 = \frac{- 4 {e}^{2 x}}{1 + {e}^{2 x}} ^ 2$

Now put it back into the derivative equation for $\cos \left(s\right)$

$\cos \left(s\right) \Rightarrow s ' \cdot - \sin \left(s\right)$

$s ' \cdot - \sin \left(s\right) = - \frac{- 4 {e}^{2 x}}{1 + {e}^{2 x}} ^ 2 \sin \left(\frac{1 - {e}^{2 x}}{1 + {e}^{2 x}}\right)$