# How do you find the derivative of  F(x) = (sinx/(1+cosx))^2?

Aug 2, 2016

$\frac{2 \sin x}{1 + \cos x} ^ 2$

#### Explanation:

The derivative of a power is stated as follows:
$\left({\left(u\right)}^{n}\right) ' = n \cdot u \cdot u '$

Applying this property the derivative of the given power is :

$\textcolor{b l u e}{\left(\sin \frac{x}{1 + \cos x} ^ 2\right) ' = 2 \left(\sin \frac{x}{1 + \cos x}\right) \left(\sin \frac{x}{1 + \cos x}\right) '}$

Let's find$\left(\sin \frac{x}{1 + \cos x}\right) '$:
The derivative of the quotient
$\left(\frac{u}{v}\right) ' = \frac{u ' v - v ' u}{v} ^ 2$

To apply the derivative of a quotient on (sinx)/(1+cosx) we've to find :

$\left(\sin x\right) ' = \cos x$
$\left(1 + \cos x\right) ' = - \sin x$
So,
$\textcolor{b l u e}{\left(\sin \frac{x}{1 + \cos x}\right) '}$
$= \frac{\left(\sin x\right) ' \left(1 + \cos x\right) - \left(1 + \cos x\right) ' \sin x}{1 + \cos x} ^ 2$
$= \frac{\cos x \left(1 + \cos x\right) - \left(- \sin x\right) \left(\sin x\right)}{1 + \cos x} ^ 2$

$= \frac{\cos x + {\left(\cos x\right)}^{2} + {\left(\sin x\right)}^{2}}{1 + \cos x} ^ 2$

Knowing that${\left(\cos x\right)}^{2} + {\left(\sin x\right)}^{2} = 1$
$= \frac{\cos x + 1}{1 + \cos x} ^ 2$

Simplifying by $1 + \cos x$
Therefore,
$\left(\sin \frac{x}{1 + \cos x}\right) ' = \frac{1}{1 + \cos x}$

So ,the derivative of the given power:
$\left({\left(\sin \frac{x}{1 + \cos x}\right)}^{2}\right) '$
$= 2 \left(\sin \frac{x}{1 + \cos x}\right) \left(\sin \frac{x}{1 + \cos x}\right) '$
$= 2 \left(\sin \frac{x}{1 + \cos x}\right) \left(\frac{1}{1 + \cos x}\right)$
$= \frac{2 \sin x}{1 + \cos x} ^ 2$