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If $y=\sin \sqrt{x}+\cos ^2 \sqrt{x}$, then $\frac{d y}{d x}$ is equal to.
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Verified Answer
$$
\text \begin{aligned}
\frac{d y}{d x} &=\frac{d}{d x} \sin \left(x^{1 / 2}\right)+\frac{d}{d x}\left[\cos \left(x^{1 / 2}\right)\right]^2 \\
&=\cos x^{1 / 2} \cdot \frac{d}{d x} x^{1 / 2}+2 \cos \left(x^{1 / 2}\right) \frac{d}{d x}\left[\cos \left(x^{1 / 2}\right)\right]
\end{aligned}
$$
$\begin{aligned}=& \cos \left(x^{1 / 2}\right) \frac{1}{2} x^{-1 / 2}+2 \cdot \cos \left(x^{1 / 2}\right) \\ &=\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}-2 \cos \left(x^{1 / 2}\right) \cdot \sin x^{1 / 2} \cdot \frac{1}{2 \sqrt{x}} \\ \because \quad & 2 \sin A \cos A=\sin 2 A \\ \therefore \quad & \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}[\cos (\sqrt{x})-\sin (2 \sqrt{x})] \end{aligned}$
\text \begin{aligned}
\frac{d y}{d x} &=\frac{d}{d x} \sin \left(x^{1 / 2}\right)+\frac{d}{d x}\left[\cos \left(x^{1 / 2}\right)\right]^2 \\
&=\cos x^{1 / 2} \cdot \frac{d}{d x} x^{1 / 2}+2 \cos \left(x^{1 / 2}\right) \frac{d}{d x}\left[\cos \left(x^{1 / 2}\right)\right]
\end{aligned}
$$
$\begin{aligned}=& \cos \left(x^{1 / 2}\right) \frac{1}{2} x^{-1 / 2}+2 \cdot \cos \left(x^{1 / 2}\right) \\ &=\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}-2 \cos \left(x^{1 / 2}\right) \cdot \sin x^{1 / 2} \cdot \frac{1}{2 \sqrt{x}} \\ \because \quad & 2 \sin A \cos A=\sin 2 A \\ \therefore \quad & \frac{d y}{d x}=\frac{1}{2 \sqrt{x}}[\cos (\sqrt{x})-\sin (2 \sqrt{x})] \end{aligned}$
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