gr8..
i came back today after so many days.. nice to find all my questions are solved :)
if f(xy)=f(x)f(y) for all x,y in R.
f'(1)=2 and f(4)=4
find f'(4)
f '(x)= [lim h->0] [f(x+h)-f(x)]/h
=[lim h -> 0] [f(x)f(1+x/h)-f(x)]/h (by substituting y=1+h/x)
=f(x)[lim h->0] [f(1+x/h)-1]/h
=[f(x)/x][lim h->0] [f(1+x/h)-1]/(h/x)
f(xy)=f(x)f(y)
putting y=4, f(4x)=4f(x)
now, x=1 f(4)=4f(1) thus, f(1)=1
using f(1)=1 in f '(x)
f '(x)=[f(x)/x] [lim h->0] [f(1+x/h)-f(1)]/(h/x)
=f(x)/x . f '(1)
=2f(x)/x
finally, x=4 gives
f '(4)=2f(4)/4=2
gr8..
i came back today after so many days.. nice to find all my questions are solved :)