Given that $f(x)=\frac{1}{x^n}$, show that $x f'(x)+n f'(x)=0$.

Question

This exercise was in my math book and of course had no solution as it's a "show" type of question. I don't see how this could hold except for when $x=-n$.

It's possible the book had a typo, and instead the question was asking you to show that $xf'(x)+nf(x)=0$ — A. S., Jun 13 '16 at 15:34

score 3 · Accepted Answer · answered Jun 13 '16 at 15:36

I think there is a typo in the problem. I am guessing the equation should actually be $$xf'(x)+nf(x)=0$$ so the second $f$ has no derivative. Assuming this, you just have to substitute. If $f(x)=\frac{1}{x^n}$, then $f'(x)=\frac{-n}{x^{n+1}}$, so $$xf'(x)+nf(x)$$ $$=\frac{-nx}{x^{n+1}}+\frac{n}{x^{n}} $$ $$=0$$

score 1 · Answer 2 · answered Jun 13 '16 at 15:30

1

Either $x=-n$ or $n=0$ are the possible solutions.

answered Jun 13 '16 at 15:30

Qwerty

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Given that $f(x)=\frac{1}{x^n}$, show that $x f'(x)+n f'(x)=0$.

2 Answers2