If \(y = \sin\sqrt{x} + \cos^{2}\sqrt{x}\), \(x \gt 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2\sqrt{x}}\left[\cos\sqrt{x} - \sin\left(2\sqrt{x}\right)\right]\)
By the chain rule, with \(\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}\):
\( \displaystyle \frac{d}{dx}\sin\sqrt{x} = \frac{\cos\sqrt{x}}{2\sqrt{x}} \)
\( \displaystyle \frac{d}{dx}\cos^{2}\sqrt{x} = 2\cos\sqrt{x} \cdot \left(-\sin\sqrt{x}\right) \cdot \frac{1}{2\sqrt{x}} \)
Using the double-angle formula \(2\sin\theta\cos\theta = \sin 2\theta\):
\( \displaystyle \frac{d}{dx}\cos^{2}\sqrt{x} = -\frac{\sin\left(2\sqrt{x}\right)}{2\sqrt{x}} \)
Adding the two:
\( \displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\left[\cos\sqrt{x} - \sin\left(2\sqrt{x}\right)\right] \)