If y = {left( {1 + frac{1}{x}} right)^x},then | vedclass

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(A) Given $y = {\left( {1 + \frac{1}{x}} \right)^x}$.Taking the natural logarithm on both sides,we get $\log y = x \log \left( {1 + \frac{1}{x}} \righ

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = {\left( {1 + \frac{1}{x}} \right)^x}$.Taking the natural logarithm on both sides,we get $\log y = x \log \left( {1 + \frac{1}{x}} \right)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left[ x \log \left( {1 + \frac{1}{x}} \right) \right]$$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log \left( {1 + \frac{1}{x}} \right) + x \cdot \frac{1}{1 + \frac{1}{x}} \cdot \left( -\frac{1}{x^2} \right)$$\frac{1}{y} \frac{dy}{dx} = \log \left( {1 + \frac{1}{x}} \right) + x \cdot \frac{x}{x + 1} \cdot \left( -\frac{1}{x^2} \right)$$\frac{1}{y} \frac{dy}{dx} = \log \left( {1 + \frac{1}{x}} \right) - \frac{1}{x + 1}$Therefore,$\frac{dy}{dx} = y \left[ \log \left( {1 + \frac{1}{x}} \right) - \frac{1}{x + 1} \right] = {\left( {1 + \frac{1}{x}} \right)^x}\left[ \log \left( {1 + \frac{1}{x}} \right) - \frac{1}{x + 1} \right]$.

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

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Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
Reason $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

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