2f(x)=f(xy)+f(xy)
⇒f(xy)=2f(x)−f(xy).......(i)
Put x=1
f(y)=2f(1)−f(1y)⇒f(y)=−f(1y)........(ii)(∵f(1)=0)
Put y=x
f(x2)=2f(x)−f(1)⇒f(x2)=2f(x)......(iii)
Put y=x2
f(x3)=2f(x2)−f(1x)⇒f(x3)=2f(x)+f(x) (∵ from (ii))⇒f(x3)=3f(x)
⇒f(xn)=nf(x)
Differentiating both the sides w.r.t. x,
nxn−1f′(xn)=nf′(x)
⇒f′(xn)=f′(x)xn−1
Now, 256=28=xn⇒x=2, n=8
∴f′(256)=f′(2)27 =256128 =2