If y =(1 + 1/x)^x, then dy/dx = A.(1 + 1/x)^x {log (1 + 1/x) - 1/(x + 1)} B. (1 + 1/x)^x log(1 + 1/x)

Question

If y =(1 + \(\cfrac{1}{\text x}\))^x, then \(\cfrac{dy}{d\mathrm x} \) =

A.\(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\)\(\Big\{log\Big(1+\cfrac{1}{\text x}\Big)\)\(-\cfrac{1}{\text x+1}\Big\}\)

B. \(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\)log\(\Big(1+\cfrac{1}{\text x}\Big)\)

C.\(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\)\(\Big\{log(\text x+1)-\cfrac{\text x}{\text x+1}\Big\}\)

D. \(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\)\(\Big\{log\Big(\text x+\cfrac{1}{\text x}\Big)+\cfrac{1}{\text x + 1}\Big\}\)

Answer 1

Correct option is  C.\(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\)\(\Big\{log(\text x+1)-\cfrac{\text x}{\text x+1}\Big\}\) 

Given y = log \(\Big(1+\cfrac{1}{\text x}\Big)^{\text x}\) 

Taking log both sides we get