Question

find f'(3pi/2) where f(x)=(sinx+1)^x

Answer 1

f(x) =(sinx +1)ˣ  at x =  3π/2   : f(x) = 0 
     take both sides log 
logf(x) = xlog(sinx + 1)
now differentiate with respect to x 
f'(x)/f(x) = x/(sinx +1) + log(sinx+1 ) 
f'(x) = f(x){x/(sinx +1) + log(sinx+1 )
  put x = 3π/2 
f'(x) = 0.{ 3π/2/(sin 3π/2 +1 ) + log(sin 3π/2+1) } 
f'(x) = 0

Answer 2

Is the answer = 0 ?
f(x) has x in the exponent. So all the derivatives will contain f(x) as a factor. f(3 pi/2) = 0 as sin 3pi/2 = -1.

Hence f '(3 pi/2) = 0.

See pic enclosed.
All derivatives will be 0 at x = 3 pi/2.