The given function is \( \displaystyle y = \cos (\tan \sqrt{x+1}) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
To find the derivative \(\displaystyle \frac{dy}{dx} \) for the given function, we’ll use the chain rule for differentiation.
\( \begin{align*} \frac{dy}{dx} &= \frac{d}{dx} \left[ \cos (\tan \sqrt{x+1}) \right] \\ &= \frac{d}{d\left(\tan\sqrt{x+1}\right)} \left[ \cos\left(\tan\sqrt{x+1}\right) \right] \cdot \frac{d}{dx} \left[ \tan\sqrt{x+1} \right] \\ &= -\sin\left(\tan\sqrt{x+1}\right) \cdot \frac{d}{dx} \left[ \tan\sqrt{x+1} \right] \\ &= -\sin\left(\tan\sqrt{x+1}\right) \cdot \frac{d}{d\left(\sqrt{x+1}\right)} \left[ \tan\left(\sqrt{x+1}\right) \right] \cdot \frac{d}{dx} \left[ \sqrt{x+1} \right] \\ &= -\sin\left(\tan\sqrt{x+1}\right) \cdot \sec^2\left(\sqrt{x+1}\right) \cdot \frac{d}{dx} \left[ \sqrt{x+1} \right] \\ &= -\sin\left(\tan\sqrt{x+1}\right) \cdot \sec^2\left(\sqrt{x+1}\right) \cdot \frac{1}{2\sqrt{x+1}} \cdot \frac{d}{dx} \left[ x + 1 \right] \\ &= -\sin\left(\tan\sqrt{x+1}\right) \cdot \sec^2\left(\sqrt{x+1}\right) \cdot \frac{1}{2\sqrt{x+1}} \cdot 1 \\ &= -\frac{\sin\bigl(\tan\sqrt{x+1}\bigr)} {2\sqrt{x+1} \cdot \cos^2\bigl(\sqrt{x+1}\bigr)} \\ \end{align*} \)Thus, the derivative of the function is:
\( \displaystyle \boxed{\frac{dy}{dx} = -\frac{\sin\bigl(\tan\sqrt{x+1}\bigr)} {2\sqrt{x+1} \cdot \cos^2\bigl(\sqrt{x+1}\bigr)} } \\ \)Please let me know in the comments if you find any errors in this solution.