The given function is \( \displaystyle y = \sin \left( \cos \left( \tan \sqrt{x} \right) \right) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
To find \( \displaystyle \frac{dy}{dx} \), we will apply the Chain Rule.
Let’s define intermediate functions:
\( \displaystyle u = \sqrt{x}, \quad v = \tan u, \quad w = \cos v, \quad y = \sin w \\ \)Differentiate each of the above functions:
\( \begin{align*} \frac{dy}{dw} &= \cos w \\ \frac{dw}{dv} &= -\sin v \\ \frac{dv}{du} &= \sec^2 u \\ \frac{du}{dx} &= \frac{1}{2\sqrt{x}} \\ \end{align*} \)Combine the derivatives using the chain rule and substitute values:
\( \begin{align*} \frac{dy}{dx} &= \frac{dy}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} \\ &= \cos w \cdot (-\sin v) \cdot \sec^2 u \cdot \frac{1}{2\sqrt{x}} \\ &= \cos \left( \cos \left( \tan \sqrt{x} \right) \right) \cdot \left( -\sin \left( \tan \sqrt{x} \right) \right) \cdot \sec^2 \left( \sqrt{x} \right) \cdot \frac{1}{2\sqrt{x}} \\ &= – \frac{1}{2\sqrt{x}} \cdot \cos \left( \cos \left( \tan \sqrt{x} \right) \right) \cdot \sin \left( \tan \sqrt{x} \right) \cdot \sec^2 \left( \sqrt{x} \right) \\ \end{align*} \)Thus, the derivative of the function is:
\( \boxed{ \frac{dy}{dx} = – \frac{1}{2\sqrt{x}} \cdot \cos \left( \cos \left( \tan \sqrt{x} \right) \right) \cdot \sin \left( \tan \sqrt{x} \right) \cdot \sec^2 \left( \sqrt{x} \right) } \\ \)Please let me know in the comments if you find any errors in this solution.