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Use mathematical induction to prove that the derivative of $f(x)=\sin(ax+b)$ is given by

$f^{(n)}(x)= (-1)^ka^n\sin(ax+b)$ if $n=2k$, and $(-1)^ka^n\cos(ax+b)$ if $n=2k+1$

for a number $k=0,1,2,3,...$

I have done som proofs by induction, but I seem to struggle as soon as trig functions appear.

Agustí Roig
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user71671
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  • Hints:$$f^{'}(x)=a cos(ax+b), f^{''}(x)=-asin(ax+b)=-af(x)$$ $$f^{3}(x)=-a^2cos(ax+b)=-a^2f^{'}(x)$$ – J.H. Sep 20 '13 at 10:21

2 Answers2

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You'll need to use the Chain Rule, together with the fact that $$\frac{d}{du}\sin(u)=\cos(u)$$ and $$\frac{d}{du}\cos(u)=-\sin(u).$$

Cameron Buie
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Let $f(x) = \sin(ax+b)$ then we claim for $n\in\mathbb N_0$ $$\begin{align*} f^{(2n)}(x) & = (-1)^n a^{2n} \sin(ax+b) \\ f^{(2n+1)}(x) & = (-1)^n a^{2n+1} \cos(ax+b) \end{align*}$$ The induction start for $n=0$ is obvious. Then we show for $k\in\mathbb N_0$: $$\begin{align*} f^{(2k)}(x) & = f^{(2k-1)'}(x) \\ & \stackrel{\text{ind.}}{=} = ((-1)^{k-1} a^{2k-1} \cos(ax+b))'\\ & = (-1)^k a^{2k} \sin(ax+b) & \text{and}\\ f^{(2k+1)}(x) & = f^{(2k)'}(x) \\ & \stackrel{\text{ind.}}{=} ((-1)^k a^{2k} \sin(ax+b))' \\ & = (-1)^k a^{2k+1} \cos(ax+b) \end{align*}$$ q.e.d.

AlexR
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