Question

$2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right),x,y\in R^{+}$
If $f\left(1\right)=0$ and $f^{\prime }\left(2\right)=256,$ then $f^{\prime }\left(256\right)$ is equal to

Answer 1

$2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right)$
$\Rightarrow f\left(xy\right)=2f\left(x\right)-f\left(\frac{x}{y}\right).......\left(i\right)$
Put $x=1$
$f\left(y\right)=2f\left(1\right)-f\left(\frac{1}{y}\right)\Rightarrow f\left(y\right)=-f\left(\frac{1}{y}\right)........\left(ii\right)(\because f(1)=0)$
Put $y=x$
$f\left(x^2\right)=2f\left(x\right)-f\left(1\right)\Rightarrow f\left(x^2\right)=2f\left(x\right)......\left(iii\right)$
Put $y=x^2$
$f\left(x^3\right)=2f\left(x^2\right)-f\left(\frac{1}{x}\right)\Rightarrow f\left(x^3\right)=2f\left(x\right)+f\left(x\right)(\because \text{from}(ii\left)\right)\Rightarrow f\left(x^3\right)=3f\left(x\right)$
$\Rightarrow f\left(x^n\right)=nf\left(x\right)$
Differentiating both the sides w.r.t. $x$,
$n{x}^{n-1}{f}^{\prime }\left(x^n\right)=n{f}^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(x^n\right)=\frac{f^{\prime }\left(x\right)}{x^{n-1}}$

Now, $256=2^8=x^n\Rightarrow x=2,n=8$

$\therefore f^{\prime }\left(256\right)=\frac{f^{\prime }\left(2\right)}{2^7}=\frac{256}{128}=2$