Differentiate the function $\sin\sqrt{x} + \cos^2\sqrt{x}$ with respect to $x$.

$\frac{1}{2\sqrt{x}} [\cos(\sqrt{x}) - \sin(2\sqrt{x})]$

The correct answer is Option (1) → $\frac{1}{2\sqrt{x}} [\cos(\sqrt{x}) - \sin(2\sqrt{x})]$ ##

Let $y = \sin\sqrt{x} + (\cos\sqrt{x})^2$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx}\sin(x^{1/2}) + \frac{d}{dx}[\cos(x^{1/2})]^2$

$= \cos x^{1/2} \cdot \frac{d}{dx} x^{1/2} + 2\cos(x^{1/2}) \cdot \frac{d}{dx} [\cos(x^{1/2})] \quad \left[ ∵\frac{d}{dx} \sin x = \cos x \right]$

$= \cos(x^{1/2}) \cdot \frac{1}{2}x^{-1/2} + 2 \cdot \cos(x^{1/2}) \cdot \left[ -\sin(x^{1/2}) \cdot \frac{d}{dx} x^{1/2} \right] \quad \left[ ∵\frac{d}{dx} x^n = nx^{n-1} \right]$

$= \cos\sqrt{x} \cdot \frac{1}{2\sqrt{x}} - 2\cos(x^{1/2}) \cdot \sin x^{1/2} \cdot \frac{1}{2\sqrt{x}}$

$= \frac{1}{2\sqrt{x}} [\cos(\sqrt{x}) - \sin(2\sqrt{x})] \quad [∵\sin 2x = 2 \sin x \cos x]$