If \(f(x) = 2\sqrt{\cos(x^{2})}\), where \(\cos(x^{2}) \gt 0\), then \(f'(x)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{2x\sin(x^{2})}{\sqrt{\cos(x^{2})}}\)
The chain rule works from the outside in: the square root, then cosine, then \(x^{2}\).
\( \displaystyle f'(x) = 2 \cdot \frac{1}{2\sqrt{\cos(x^{2})}} \cdot \left(-\sin(x^{2})\right) \cdot 2x \)
\( \displaystyle = -\frac{2x\sin(x^{2})}{\sqrt{\cos(x^{2})}} \)