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प्रश्न
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)]`
\[\left( 1 + \frac{1}{x} \right)^x \log \left( 1 + \frac{1}{x} \right)\]
\[\left( x + \frac{1}{x} \right)^x \left\{ \log \left( x + 1 \right) - \frac{x}{x + 1} \right\}\]
\[\left( x + \frac{1}{x} \right)^x \left\{ \log \left( 1 + \frac{1}{x} \right) + \frac{1}{x + 1} \right\}\]
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उत्तर
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)])`.
\[\text{Let y }= \left( 1 + \frac{1}{x} \right)^x \]
\[\text{ Taking log on both sides}, \]
\[\log y = x \log\left( 1 + \frac{1}{x} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\frac{d}{dx}\log\left( 1 + \frac{1}{x}\right) + \log\left( 1 + \frac{1}{x} \right)\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x\left( \frac{1}{1 + \frac{1}{x}}\right)\frac{d}{dx}\left( 1 + \frac{1}{x} \right) + \log\left( 1 + \frac{1}{x} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = x \times \frac{x}{x + 1}\left( - \frac{1}{x^2} \right) + \log\left( 1 + \frac{1}{x} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{x^2}{x + 1} \times \frac{- 1}{x^2} + \log\left( 1 + \frac{1}{x} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{- 1}{x + 1} + \log\left( 1 + \frac{1}{x} \right)\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \frac{- 1}{x + 1} + \log\left( 1 + \frac{1}{x} \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = \left( 1 + \frac{1}{x} \right)^x \left[ \log\left( 1 + \frac{1}{x} \right) - \frac{1}{x + 1} \right]\]
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