# How do you find the derivative of sinx/(1+cosx)?

May 3, 2018

$\frac{1}{\cos x + 1}$

#### Explanation:

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

$f ' \left(x\right) = \left(\sin \frac{x}{\cos x + 1}\right) '$

The derivative of $f \frac{x}{g} \left(x\right)$ using Quotient Rule is
$\frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} ^ 2 \left(x\right)$
so in our case it is

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

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

$\frac{\textcolor{b l u e}{{\cos}^{2} x} + \cos x + \textcolor{b l u e}{{\sin}^{2} x}}{\cos x + 1} ^ 2$ $=$

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

$\frac{1}{\cos x + 1}$

May 3, 2018

$\frac{1}{2} {\sec}^{2} \left(\frac{x}{2}\right) \mathmr{and} \frac{1}{1 + \cos x}$.

#### Explanation:

We have, $\sin \frac{x}{1 + \cos x}$,

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

$= \tan \left(\frac{x}{2}\right)$.

$\text{Therefore, } \frac{d}{\mathrm{dx}} \left\{\sin \frac{x}{1 + \cos x}\right\}$,

$= \frac{d}{\mathrm{dx}} \left\{\tan \left(\frac{x}{2}\right)\right\}$,

$= {\sec}^{2} \left(\frac{x}{2}\right) \cdot \frac{d}{\mathrm{dx}} \left\{\frac{x}{2}\right\} \ldots \ldots \text{[The Chain Rule]}$,

$= {\sec}^{2} \left(\frac{x}{2}\right) \cdot \frac{1}{2}$,

$= \frac{1}{2} {\sec}^{2} \left(\frac{x}{2}\right) , \mathmr{and} ,$

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

$= \frac{1}{1 + \cos x}$.