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Can someone help me prove that the $n$-th derivative of $f(x)=x\cos(x)$ is:

$$f^n(x)=x\cos\left(x+\frac{n\pi}{2}\right)+n\cos\left(x+(n-1)\frac{\pi}{2}\right)?$$

Thanks, and stay safe.

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1 Answer

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Just using induction using the product rule:

$f(x) = x\cos x$ so

$f'(x) = \cos x + x(-\sin x) = \cos x - x\sin x$

$f^2(x) = -\sin x - \sin x - x\cos x= -2\sin x - x\cos x$

$f^3(x) = -2\cos x-\cos x - x(-\sin x) = -3\sin x +x\sin x$.

By pattern recognition we sort of see that $f^n = \pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x$. Bearing in mind that $\sin x = \cos (\frac \pi 2 - x)$ and that the pattern of the $\pm$ has a periodicity of $4$ and so on we can see that

$f^n(x) =\pm n \operatorname{trig} x \pm x\operatorname{other\ trig} x=$

$x\cos(x+\frac{n\pi}{2})+n\cos(x+(n-1)\frac{\pi}{2})$

and we can feel grateful that someone else hammered the details and fine polished it for us.

But we have to verify it.

For $n= 0$ this fits

$f(x) = x\cos x = x\cos(x+\frac{0\pi}{2})+0\cos(x+(0-1)\frac{\pi}{2})$

No question of that.

And for $n=1$ we have

$f'(x) = \cos x - x\sin x= x\cos(x+\frac{1\pi}{2})+1\cos(x+(1-1)\frac{\pi}{2})$

$=1\cos (x + 0) + x\cos(x+\frac {\pi}2) = \cos x- x \sin x$.

It works. Now to verify in general we must use induction.

If $f^k(x) =x\cos(x+\frac{k\pi}{2})+k\cos(x+(k-1)\frac{\pi}{2})$

then

$f^{k+1}(x) = \cos(x+\frac{k\pi}{2}) + x(-\sin(x+\frac{k\pi}2) -k\sin(x+(k-1)\frac \pi 2)=$

$=\cos(x+\frac{k\pi}{2}) +k\cos(x+k\frac {\pi}2) + x(\cos(x+\frac {(k+1)\pi}2))=$

$x\cos (x+\frac {(k+1)\pi}2)+ (k+1)\cos(x+k\frac \pi 2)$

So the result follows by induction.

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