3. find the derivative of f(x) = 1 / (sin(x - sinx))

Explanation:

Step1: Rewrite the function

We can rewrite $f(x)=\frac{1}{\sin x-\sin^{2}x}=(\sin x - \sin^{2}x)^{-1}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{-1}$ and $u=\sin x-\sin^{2}x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: If $y = u^{-1}$, then $\frac{dy}{du}=-u^{-2}=-\frac{1}{u^{2}}$. Second, find $\frac{du}{dx}$: Since $u=\sin x-\sin^{2}x$, $\frac{du}{dx}=\cos x - 2\sin x\cos x=\cos x(1 - 2\sin x)$.

Step3: Calculate the derivative of $f(x)$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-\frac{1}{(\sin x-\sin^{2}x)^{2}}\cdot\cos x(1 - 2\sin x)=\frac{\cos x(2\sin x - 1)}{(\sin x-\sin^{2}x)^{2}}$.

Answer:

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