find the derivative of the following function.\ny = \\frac{7x\\sin x}{1 - \\cos x}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 7x\sin x$ and $v=1 - \cos x$.

Step2: Find $u^\prime$ using product - rule

The product - rule states that if $u = f(x)g(x)$ where $f(x)=7x$ and $g(x)=\sin x$, then $u^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)$. Since $f^\prime(x)=7$ and $g^\prime(x)=\cos x$, we have $u^\prime=7\sin x + 7x\cos x$. And $v^\prime=\sin x$.

Step3: Substitute $u$, $u^\prime$, $v$, $v^\prime$ into quotient - rule

[ \begin{align*} \frac{dy}{dx}&=\frac{(7\sin x + 7x\cos x)(1 - \cos x)-7x\sin x(\sin x)}{(1 - \cos x)^{2}}\ &=\frac{7\sin x-7\sin x\cos x + 7x\cos x-7x\cos^{2}x-7x\sin^{2}x}{(1 - \cos x)^{2}}\ &=\frac{7\sin x-7\sin x\cos x + 7x\cos x-7x(\sin^{2}x+\cos^{2}x)}{(1 - \cos x)^{2}} \end{align*} ] Since $\sin^{2}x+\cos^{2}x = 1$, we get: [ \begin{align*} \frac{dy}{dx}&=\frac{7\sin x-7\sin x\cos x + 7x\cos x-7x}{(1 - \cos x)^{2}} \end{align*} ]

Answer:

$\frac{7\sin x-7\sin x\cos x + 7x\cos x-7x}{(1 - \cos x)^{2}}$