If $y=(1+x)^{1 / x}$, then $\frac{d y}{d x}=$

$\frac{x}{1+x}(1+x)^{1 / x}-\log (1+x)$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y=(1+x)^{1 / x}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: B

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

Option B (

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

) is correct because after applying logarithmic differentiation and simplifying, the derivative is found to be

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

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