f(xy)=f(x)f(y) ......(1) Put x=y=0 in (1) to getf(0)=1 Put x=y=1 in (1) to get f(1)=0 or f(1)=1 f(1)=0 is rejected else y=1 in (1) gives f(x)=0 imply f(0)=0 Hence, f(0)=1 and f(1)=1 By first principle derivative formula, f′(x)=
lim
h→0
f(x+h)−f(x)
h
lim
h→0
f(x)(
f(1+
h
x
)−f(1)
h
) ⇒f′(x)=
f(x)
x
f′(1) ⇒
f′(x)
f(x)
=
k
x
⇒lnf(x)=klnx+c f(1)=1⇒ln1=kln1+c⇒c=0 ⇒lnf(x)=klnx⇒f(x)=xk but f(0)=1 ⇒k=0 ∴f(x)=1