Question

Let $f\left(x^{3} + y^{3}\right)=xf\left(x^{2}\right)+yf\left(y^{2}\right)$ for all $x,y\in ℝ$ and $v$ is differentiable. If $f^{'}\left(0\right)=5,$ then find value of $f^{'}\left(5\right)$

Answer 1

Let $f\left(x^{3} + y^{3}\right)=xf\left(x^{2}\right)+yf\left(y^{2}\right)$ for all
$x,y\in R$ and $f\left(x\right)$ is differentiable.
If $f^{'}\left(0\right)=5,$ then find value of $f^{'}\left(5\right)$
$x=y=0\Rightarrow f\left(0\right)=0$
$y=0\Rightarrow f\left(x^{3}\right)=xf\left(x^{2}\right)$
$\Rightarrow f\left(x\right)=cx$ for some constant $C$
So, $f^{'}\left(x\right)=C\Rightarrow f^{'}\left(0\right)=f^{'}\left(5\right)=5$