If y=sqrt{e^{sqrt{x}}},then frac{d y}{d x}= | vedclass

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(D) Given $y = \sqrt{e^{\sqrt{x}}}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{\sqrt{x}})^{1/2} = \frac{1}{2} \sqrt{x} \ln e = \frac{\

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$\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}$

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(D) Given $y = \sqrt{e^{\sqrt{x}}}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{\sqrt{x}})^{1/2} = \frac{1}{2} \sqrt{x} \ln e = \frac{\sqrt{x}}{2}$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{4\sqrt{x}}$.Therefore,$\frac{dy}{dx} = \frac{y}{4\sqrt{x}}$.Substituting $y = \sqrt{e^{\sqrt{x}}}$:$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

If $y=\sqrt{e^{\sqrt{x}}}$,then $\frac{d y}{d x}=$

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