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# Solution for If Y = ( 1 + 1 X ) X , Then D Y D X = (A) ( 1 + 1 X ) X ( 1 + 1 X ) − 1 X + 1 - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

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

(a) $\left( 1 + \frac{1}{x} \right)^x \left( 1 + \frac{1}{x} \right) - \frac{1}{x + 1}$

(b) $\left( 1 + \frac{1}{x} \right)^x \log \left( 1 + \frac{1}{x} \right)$

(c) $\left( x + \frac{1}{x} \right)^x \left\{ \log \left( x + 1 \right) - \frac{x}{x + 1} \right\}$

(d) $\left( x + \frac{1}{x} \right)^x \left\{ \log \left( 1 + \frac{1}{x} \right) + \frac{1}{x + 1} \right\}$

#### Solution

(a)  $\left( 1 + \frac{1}{x} \right)^x \left( 1 + \frac{1}{x} \right) - \frac{1}{x + 1}$

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

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Solution If Y = ( 1 + 1 X ) X , Then D Y D X = (A) ( 1 + 1 X ) X ( 1 + 1 X ) − 1 X + 1 Concept: Simple Problems on Applications of Derivatives.
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