# 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!

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Have you studied the chain rule? –  hardmath Oct 27 '12 at 20:58
$\sin^n \theta$ is a more convenient way of writing $[\sin \theta]^n$. –  BobaFret Oct 27 '12 at 21:04
@BobaFret: Ok thanks, I'll remember. –  calbertts Oct 27 '12 at 21:07

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

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Your answer is very nice!, thanks –  calbertts Oct 27 '12 at 21:14
"first you take the derivative of $\sin^{n-1}(x)$"...with respect to what? –  Cameron Buie Oct 29 '12 at 2:55

$\sin^{n-1}\theta$ 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: $(f\circ g)'(\theta)=g'(\theta)f'(g(\theta))$.

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Thanks by your time, now I understand. –  calbertts Oct 27 '12 at 21:13

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, $$\frac{d}{d\theta}\left[\sin^{n-1}\theta\right]=\frac{d}{d\theta}\left[u^{n-1}\right]=\frac{d}{du}\left[u^{n-1}\right]\cdot\frac{du}{d\theta}=(n-1)\sin^{n-2}\theta\cdot\cos\theta.$$

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Thanks by your time. –  calbertts Oct 27 '12 at 21:09

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 $\cos(x)$.

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Thanks by your time. –  calbertts Oct 27 '12 at 21:12

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 $$=\cos x \sin ^{n-1} x +\sin x (n-1) \cos x \sin ^{n-2} x = n \cos x \sin ^{n-1} x$$

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Good!, thanks so much! –  calbertts Oct 27 '12 at 21:29