JEE Main & Advanced Mathematics Differentiation Question Bank Differentiation of implicit function Parametric

  • question_answer
    If  \[y={{\left( 1+\frac{1}{x} \right)}^{x}}\], then \[\frac{dy}{dx}=\]                                         [BIT  Ranchi 1992]

    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)            \[{{\left( x+\frac{1}{x} \right)}^{x}}\left[ \log (x-1)-\frac{x}{x+1} \right]\]

    D)            \[{{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)+\frac{1}{1+x} \right]\]

    Correct Answer: A

    Solution :

               \[y={{\left( 1+\frac{1}{x} \right)}^{x}}\Rightarrow \log y=x\log \left( 1+\frac{1}{x} \right)\]                    \[\Rightarrow \frac{1}{y}\frac{dy}{dx}=\log \left( 1+\frac{1}{x} \right)-\frac{1}{1+x}\]                    Þ \[\frac{dy}{dx}={{\left( 1+\frac{1}{x} \right)}^{x}}\left[ \log \left( 1+\frac{1}{x} \right)-\frac{1}{1+x} \right]\].


You need to login to perform this action.
You will be redirected in 3 sec spinner