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

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Aicharya · Accepted Answer

$\begin{aligned} & 2 \log y=\sqrt{x} \log e \Rightarrow 2 \log y=\sqrt{x} \ & \frac{2}{y} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{4 \sqrt{x}}ight]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}} \end{aligned}$ Taking $\log$ on both sides, $2 \log \mathrm{y}=\sqrt{\mathrm{x}} \log \mathrm{e} \Rightarrow 2 \log \mathrm{y}=\sqrt{\mathrm{x}}$ Differentiating both sides w.r.t. $x$, we get $\frac{2}{y} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{4 \sqrt{x}}ight]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}$

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

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