A) \[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)-\frac{1}{1+x} \right]\]
B) \[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right) \right]\]
C) \[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log (x+1)-\frac{x}{x+1} \right]\]
D) \[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)+\frac{x}{1+x} \right]\]
Correct Answer: A
Solution :
Given, \[y={{\left( 1+\frac{1}{x} \right)}^{x}}\] Taking log on both sides, we get \[\log y=x\log \left( 1+\frac{1}{x} \right)\] On differentiating w.r.t.\[x,\]we get \[\frac{1}{y}\frac{dy}{dx}=x.\frac{1}{\left( 1+\frac{1}{x} \right)}\left( 0-\frac{1}{{{x}^{2}}} \right)+\log \left( 1+\frac{1}{x} \right)\] \[\Rightarrow \] \[\frac{1}{y}\frac{dy}{dx}=\frac{x.x}{(x+1)}\left( \frac{-1}{{{x}^{2}}} \right)+\log \left( 1+\frac{1}{x} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=y\left[ \log \left( 1+\frac{1}{x} \right)-\frac{1}{x+1} \right]\] \[\Rightarrow \] \[\frac{dy}{dx}={{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)-\frac{1}{x+1} \right]\]
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