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

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Question

(C) Given $y = (1 + x)^x$. Taking the natural logarithm on both sides,we get $\log y = x \log(1 + x)$. Differentiating both sides with respect to $x$

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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