Question

If \[y = \left( 1 + \frac{1}{x} \right)^x , \text{then} \frac{dy}{dx} =\] ____________.

विकल्प

• `(1+1/x)^x [log (1+1/x)-1/(1+x)]`

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

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

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

Answer 1

If \[y = \left( 1 + \frac{1}{x} \right)^x , \text{then} \frac{dy}{dx} =\] `\underline((1+1/x)^x [log (1+1/x)-1/(1+x)])`.

\[\text{Let y }= \left( 1 + \frac{1}{x} \right)^x \]

\[\text{ Taking log on both sides}, \]

\[\log y = x \log\left( 1 + \frac{1}{x} \right)\]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\log\left( 1 + \frac{1}{x}\right) + \log\left( 1 + \frac{1}{x} \right)\frac{d}{dx}\left( x \right) \]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left( \frac{1}{1 + \frac{1}{x}}\right)\frac{d}{dx}\left( 1 + \frac{1}{x} \right) + \log\left( 1 + \frac{1}{x} \right)\]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x \times \frac{x}{x + 1}\left( - \frac{1}{x^2} \right) + \log\left( 1 + \frac{1}{x} \right)\]

\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{x^2}{x + 1} \times \frac{- 1}{x^2} + \log\left( 1 + \frac{1}{x} \right)\]

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

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

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