f(xy)=e^(xy-x-y)[e^y.f(x)+e^x.f(y)]

f(1)=2e^-1  (ef(1))

so f(1)=0

 

this is how we do

f`(x)=_{[h ][0 ]} [f(x+h)-f(x)]/h       -------------------   1

now we will write
 

f(x(1+h/x))=e^h(1-1/x)e^-1 [e^1+h/x.f(x)+e^x.f(1+h/x)]

 

 

using the equation 1 we get!!!
f`(x)=_{[h ][0 ]} f(x)(e^h-1)/h+e^x-1{f(1+h/x)-f(1)}/(h/x)}/x

now take the limit of h tending to zero we get

f`(x)=f(x)+(e<sup>x-1</sup>/x)*f`(1)

f`(1)=e

so we get

f`(x)=f(x)+(e^x-1/x)*e

f`(x)=f(x)+(e^x/x)

 

Now this is a linear differencial equation with I.F =e^-x

 

on solving we get

 

f(x)

e^-xf(x)=logx+C

f(1)=0

so C=0

f(x)=logx*e^x

 

 

 

 

bye !!!!!