If y=x^{sqrt{x}},then frac{dy}{dx}= | vedclass

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(D) Given,$y=x^{\sqrt{x}}$.Taking $\ln$ on both sides,we get $\ln y = \sqrt{x} \ln x$.Differentiating both sides with respect to $x$ using the product

Vedclass · Accepted Answer

$\frac{y(\ln x+2)}{2 \sqrt{x}}$

Vedclass · Answer

(D) Given,$y=x^{\sqrt{x}}$.Taking $\ln$ on both sides,we get $\ln y = \sqrt{x} \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \sqrt{x} \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(\sqrt{x})$.$\frac{1}{y} \frac{dy}{dx} = \sqrt{x} \cdot \frac{1}{x} + \ln x \cdot \frac{1}{2\sqrt{x}}$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}}$.$\frac{1}{y} \frac{dy}{dx} = \frac{2 + \ln x}{2\sqrt{x}}$.Therefore,$\frac{dy}{dx} = \frac{y(2 + \ln x)}{2\sqrt{x}}$.Thus,option $D$ is correct.

If $y=x^{\sqrt{x}}$,then $\frac{dy}{dx}=$

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