Derivate of $(\sin \theta)^{n-1}$ Please, explain me why the derivative of $$(\sin \theta)^{n-1} = (n-1)\cos\theta \cdot (\sin \theta)^{n-2}$$
Thanks so much!
 A: $\sin^{n-1}\theta$ is the composition $f\circ g$ of $f\colon x\mapsto x^{n-1}$ and $g\colon x\mapsto \sin x$. Therefore the chain rule applies: $(f\circ g)'(\theta)=g'(\theta)f'(g(\theta))$.
A: Using chain rule, we get that $$\dfrac{d ((f(x))^k)}{dx} = \dfrac{d ((f(x))^k)}{d f(x)} \dfrac{d f(x)}{dx} = k (f(x))^{k-1} \dfrac{df(x)}{dx}$$
In you case, $f(x) = \sin(x)$, $k=n-1$. Recall that the derivative of $\sin(x)$ is $\cos(x)$.
A: A combination of power rule and chain rule.
Put $u=\sin\theta$, so that $\sin^{n-1}\theta=u^{n-1}$. By power rule, $$\frac{d}{du}\left[u^{n-1}\right]=(n-1)u^{n-2}=(n-1)\sin^{n-2}\theta.$$ By chain rule, $$\frac{d}{d\theta}\left[\sin^{n-1}\theta\right]=\frac{d}{d\theta}\left[u^{n-1}\right]=\frac{d}{du}\left[u^{n-1}\right]\cdot\frac{du}{d\theta}=(n-1)\sin^{n-2}\theta\cdot\cos\theta.$$
A: You can prove it by induction and using the product rule of differentiation
i.e. $(f\cdot g)'= f'\cdot g+f\cdot g'$
for $k=2$
it holds $(\sin ^{2-1}x)'= (\sin x )'= (2-1)\cos x \sin ^0 x$
assume it holds for 
$k=n-1$
$$(\sin ^n x )'=(\sin x \sin ^{n-1} x)'= (\sin x)'\sin ^{n-1} x +\sin x (\sin ^{n-1} x )'$$
by induction we get
$$=\cos x \sin ^{n-1} x +\sin x (n-1) \cos x \sin ^{n-2} x = n \cos x \sin ^{n-1} x$$
A: It's chain rule:
first you take the derivative of $\sin^{n-1} (x)$, it becomes $(n-1)\sin^{n-2} (x)$ then you have to take the derivative of $\sin( x)$ which is $\cos (x)$ and so the answer is $(n-1)\cos (x)  \sin^{n-2} (x)$
