Question

The function $f:R\to R$ satisfies $f\left(x^2\right).f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right).f^{\prime }\left(x^2\right)$ for all real $x$. Given that $f\left(1\right)=1$ and compute the value of $f^{\prime }\left(1\right)+f^{\prime \prime }\left(1\right)$.

Answer 1

Given : $f\left(x^2\right).f^{{}^{\prime \prime }}\left(x\right)=f^{{}^{\prime }}\left(x\right).f^{{}^{\prime }}\left(x^2\right)\forall x\to \left(1\right)$ and $f\left(1\right)=1$

Since $f\left(1\right)=1$

Assumptions :

Let $f\left(x\right)=\frac{1}{x}sinceheref\left(1\right)=1$

$f^{{}^{\prime }}\left(x\right)=\frac{-1}{x^2}\Rightarrow f^{{}^{\prime }}\left(1\right)=\frac{-1}{1}=-1$

$f^{{}^{\prime \prime }}\left(x\right)=\frac{+1}{x^3}\Rightarrow f^{{}^{\prime }}\left(1\right)=\frac{1}{1}=-1$

$f\left(x^2\right)=\frac{1}{x^2}$

$f^{{}^{\prime }}\left(x^2\right)=\frac{-1}{x^3}$

$\Rightarrow f\left(x^2\right).f^{{}^{\prime \prime }}\left(x\right)=f^{{}^{\prime }}\left(x\right).f^{{}^{\prime }}\left(x^2\right)$

$\Rightarrow \frac{1}{x^2}.\frac{1}{x^3}=\frac{-1}{x^2}.\frac{-1}{x^3}$

$\Rightarrow \frac{1}{x^5}=\frac{1}{x^5}$

Therefore , $\Rightarrow f\left(x\right)=\frac{1}{x}$

$\therefore f^{{}^{\prime }}\left(1\right)+f^{{}^{\prime \prime }}\left(1\right)\Rightarrow -1+1\Rightarrow 0$

Hence the answer is zero [0].