Question

Let $f\left(x\right)=x^n,n\in N$, then the value of $n,$ for which $f^{\prime }(a+b)=f^{\prime }\left(a\right)+f^{\prime }\left(b\right)$ is valid for $a,b>0,$ is equal to

Answer 1

$f\left(x\right)=x^n$ $\Rightarrow f^{\prime }\left(x\right)=n{x}^{n-1}$
$f^{\prime }(a+b)=f^{\prime }\left(a\right)+f^{\prime }\left(b\right)$
$\Rightarrow n(a+b{)}^{n-1}=n{a}^{n-1}+n{b}^{n-1}$
$\Rightarrow (a+b{)}^{n-1}=a^{n-1}+b^{n-1}$
which is possible iff $n-1=1\Rightarrow n=2$