If \(y = \cos\left(\tan\sqrt{x + 1}\right)\), \(x \gt -1\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{\sin\left(\tan\sqrt{x + 1}\right)\sec^{2}\sqrt{x + 1}}{2\sqrt{x + 1}}\)
The chain rule works from the outside in: cosine, then tangent, then the square root of x + 1.
\( \displaystyle \frac{d}{dx}\sqrt{x + 1} = \frac{1}{2\sqrt{x + 1}} \)
\( \displaystyle \frac{d}{dx}\tan\sqrt{x + 1} = \sec^{2}\sqrt{x + 1} \cdot \frac{1}{2\sqrt{x + 1}} \)
\( \displaystyle \frac{dy}{dx} = -\sin\left(\tan\sqrt{x + 1}\right) \cdot \frac{\sec^{2}\sqrt{x + 1}}{2\sqrt{x + 1}} \)