If $f(x)=x^n, n$ being a non-negative integer, then the values of $n$ for which $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$, for all $\alpha, \beta>0$, are

0, 2

We have,

$f(x)=x^n$

$\Rightarrow f^{\prime}(x)=n x^{n-1}$

∴  $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta>0$

$\Rightarrow n(\alpha+\beta)^{n-1}=n \alpha^{n-1}+n \beta^{n-1}$ for all $\alpha, \beta>0$

$\Rightarrow (\alpha+\beta)^{n-1}=\alpha^{n-1}+\beta^{n-1}$ for all $\alpha, \beta>0$

$\Rightarrow n-1=1 \Rightarrow n=2$

Also, for n = 0, we have

f(x) = 1  for all  x

$\Rightarrow f^{\prime}(x)=0$ for all  x

∴  $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$

Hence, n = 0, 2.