Differentiation Question 455
Question: If $ y={{( 1+\frac{1}{x} )}^{x}} $ , then $ \frac{dy}{dx}= $
[BIT Ranchi 1992]
Options:
A) $ {{( 1+\frac{1}{x} )}^{x}}[ \log ( 1+\frac{1}{x} )-\frac{1}{1+x} ] $
B) $ {{( 1+\frac{1}{x} )}^{x}}[ \log ( 1+\frac{1}{x} ) ] $
C) $ {{( x+\frac{1}{x} )}^{x}}[ \log (x-1)-\frac{x}{x+1} ] $
D) $ {{( 1+\frac{1}{x} )}^{x}}[ \log ( 1+\frac{1}{x} )+\frac{1}{1+x} ] $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{( 1+\frac{1}{x} )}^{x}}\Rightarrow \log y=x\log ( 1+\frac{1}{x} ) $
$ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\log ( 1+\frac{1}{x} )-\frac{1}{1+x} $
Therefore $ \frac{dy}{dx}={{( 1+\frac{1}{x} )}^{x}}[ \log ( 1+\frac{1}{x} )-\frac{1}{1+x} ] $ .
BETA
