Question

find the derivative of the following function. y = \frac{7x\sin x}{1 - \cos x} \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 $u^\prime$

Use the product - rule $(fg)^\prime=f^\prime g+fg^\prime$. If $f = 7x$ and $g=\sin x$, then $f^\prime=7$ and $g^\prime=\cos x$. So, $u^\prime=7\sin x+7x\cos x$.

Step3: Find $v^\prime$

The derivative of $v = 1-\cos x$ is $v^\prime=\sin x$.

Step4: Substitute $u$, $u^\prime$, $v$ and $v^\prime$ into quotient - rule

$y^\prime=\frac{(7\sin x + 7x\cos x)(1 - \cos x)-7x\sin x(\sin x)}{(1 - \cos x)^{2}}$.
Expand the numerator:
\[

$$\begin{align*} &(7\sin x + 7x\cos x)(1 - \cos x)-7x\sin^{2}x\\ =&7\sin x-7\sin x\cos x+7x\cos x - 7x\cos^{2}x-7x\sin^{2}x\\ =&7\sin x-7\sin x\cos x+7x\cos x - 7x(\cos^{2}x+\sin^{2}x) \end{align*}$$

\]
Since $\cos^{2}x+\sin^{2}x = 1$, the numerator becomes $7\sin x-7\sin x\cos x+7x\cos x - 7x$.
So, $y^\prime=\frac{7\sin x-7\sin x\cos x+7x\cos x - 7x}{(1 - \cos x)^{2}}$.

Answer:

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