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If $y=\sqrt{e^{\sqrt{x}}}$, then $\frac{d y}{d x}=$
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1397 Upvotes
Verified Answer
The correct answer is:
$\frac{e^{\frac{\sqrt{x}}{2}}}{4 \sqrt{x}}$
$$
\begin{aligned}
& 2 \log y=\sqrt{x} \log e \Rightarrow 2 \log y=\sqrt{x} \\
& \frac{2}{y} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{4 \sqrt{x}}\right]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}
\end{aligned}
$$
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
$$
\frac{2}{y} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{4 \sqrt{x}}\right]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}
$$
\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}}\right]=\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}}\right]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}
$$
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