Question

find $\frac{dy}{dx}$ for $y = \frac{cos x}{1+cos x}$.
$\frac{dy}{dx}=square$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = \cos x$, $u'=-\sin x$, $v = 1+\cos x$, and $v'=-\sin x$.

Step2: Substitute values into quotient - rule

$\frac{dy}{dx}=\frac{(-\sin x)(1 + \cos x)-\cos x(-\sin x)}{(1 + \cos x)^{2}}$.

Step3: Expand the numerator

$(-\sin x)(1 + \cos x)-\cos x(-\sin x)=-\sin x-\sin x\cos x+\sin x\cos x$.

Step4: Simplify the numerator

$-\sin x-\sin x\cos x+\sin x\cos x=-\sin x$.

Answer:

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