if the function satisfies the relation f(x+y)=y|x-1|/(x-1)f(x)+f(y) with f(1)=2, then lim f`(x) where x tends to 1 is

f(x+y)=[y|x-1| ]/ [(x-1)f(x)+f(y)]
We take log on both sides and have
Log f(x+y) = log y + log mod x-1 –log (x-1) f(x) +f(y0
We find derivatie now
F’(x+y)(1+y’)/f(x+y) = y’/y + 1/mod x-1 -1/x-1 f(x) + f(y)
When x>1 1/x-1 absolute value = 1/x-1.
When x<1 we cannot find as x-1 becomes negative and log does not exist for negative numbers.
As here y is also there, I am unable to arrive at the exact answer for f’(x)
If it single variable easily we can find out
So what we do is we put x=1 and y =0 and try that
We get f(1) =2 which means 2
Regards,
Nirmal Singh
Askiitians Faculty