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 $f^{{}^{\prime }}\left(0\right)=5,$ then find value of $f^{{}^{\prime }}\left(5\right)$

Answer 1

Answer: 5

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 $f^{{}^{\prime }}\left(0\right)=5,$ then find value of $f^{{}^{\prime }}\left(5\right)$
$x=y=0\Rightarrow f\left(0\right)=0$
$y=0\Rightarrow f\left(x^3\right)=xf\left(x^2\right)$
$\Rightarrow f\left(x\right)=cx$ for some constant $C$
So, $f^{{}^{\prime }}\left(x\right)=C\Rightarrow f^{{}^{\prime }}\left(0\right)=f^{{}^{\prime }}\left(5\right)=5$