derivative of (am/(am+n))^m wrt m

Question

My guess is

 

y=(am/(am+n))^m

F.O.C wrt m

 

lny = m* ln(am/(am+n)) (ln on both sides)

 

1/y * y'= ln(am/(am+n) + m*((am+n)/am)*(a(am+n)-am(a))/(am+n)^2 (chain rule on left hand side, product rule and chain rule on right hand side)

 

y'= y*{ln(am/(am+n))+n/((am+n))}

  = (am/(am+n))^m*{ln(am/(am+n))+n/((am+n))}

 

I am not sure if I have any mistakes in my logic or not.

Could anyone tell me if this is correct or not?

Answer 1

y=(am/(am+n))^m 

log(y)=mlog(am)-mlog(am+n)

y'/y=log(am)+a-log(am+n)-a=log(am)-log(am+n)=log(am/(am+n))

y'=ylog(am/(am+n))=(am/(am+n))^m*log(am/(am+n))