find $\frac{dy}{dx}$ for $y = \frac{cos x}{1+cos x}$.\n$\frac{dy}{dx}=square$
find $\frac{dy}{dx}$ for $y = \frac{cos x}{1+cos x}$.\n$\frac{dy}{dx}=square$
Answer
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:
$\frac{-\sin x}{(1 + \cos x)^{2}}$