Question

A function $f\left(x\right)$ is so defined that for all $x$, ${\left[f\left(x\right)\right]}^n=f\left(nx\right)$. If $f^{\prime }\left(x\right)$ denotes derivative of $f\left(x\right)$ w.r.t $x$, then show that $f^{\prime }\left(x\right).f\left(nx\right)=f\left(x\right).f^{\prime }\left(nx\right)$.

Answer 1

We have, ${\left[f\left(x\right)\right]}^n=f\left(nx\right)$

Differentiating w.r.t. $x,$ we get

$n{\left[f\left(x\right)\right]}^{n-1}.f^{\prime }\left(x\right)=f^{\prime }\left(nx\right).n$

$\Rightarrow {\left[f\left(x\right)\right]}^{n-1}.f^{\prime }\left(x\right)=f^{\prime }\left(nx\right)$

$\Rightarrow {\left[f\left(x\right)\right]}^n.f^{\prime }\left(x\right)=f^{\prime }\left(nx\right).f\left(x\right)$

$[$Multiplying both sides by $f\left(x\right)]$

$\Rightarrow f\left(nx\right).f^{\prime }\left(x\right)=f^{\prime }\left(nx\right).f\left(x\right)$ $[\because {\left[f\left(x\right)\right]}^n=f\left(nx\right)]$