The given function is \( y = \sin \sqrt{\cos \sqrt{\tan m x}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
To find the derivative of the function \( y = \sin \sqrt{\cos \sqrt{\tan m x}} \) with respect to \( x \), we apply the chain rule repeatedly:
\( \displaystyle \frac{dy}{dx} = \frac{d}{dx} \left( \sin \sqrt{\cos \sqrt{\tan m x}} \right) \\ \displaystyle = \frac{d \left( \sin \sqrt{\cos \sqrt{\tan m x}} \right)}{d\left( \sqrt{\cos \sqrt{\tan m x}} \right)} \cdot \frac{d \left(\sqrt{\cos \sqrt{\tan m x}} \right)}{dx} \\ \displaystyle = \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \frac{1}{2 \sqrt{\cos \sqrt{\tan m x}}} \cdot \frac{d}{dx} \left( \cos \sqrt{\tan m x} \right) \\ \displaystyle = \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \frac{1}{2 \sqrt{\cos \sqrt{\tan m x}}} \cdot -\sin \sqrt{\tan m x} \cdot \frac{d}{dx} \left( \sqrt{\tan m x} \right) \\ \displaystyle = – \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \frac{1}{2 \sqrt{\cos \sqrt{\tan m x}}} \cdot \sin \sqrt{\tan m x} \cdot \frac{1}{2 \sqrt{\tan m x}} \cdot \frac{d}{dx} (\tan m x) \\ \displaystyle = – \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \frac{1}{2 \sqrt{\cos \sqrt{\tan m x}}} \cdot \sin \sqrt{\tan m x} \cdot \frac{1}{2 \sqrt{\tan m x}} \cdot m \sec^2 m x \\ \displaystyle = – \frac{m \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \sin \sqrt{\tan m x} \cdot \sec^2 m x}{4 \sqrt{\cos \sqrt{\tan m x}} \cdot \sqrt{\tan m x}} \\ \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = -\frac{m \cos \sqrt{\cos \sqrt{\tan m x}} \cdot \sin \sqrt{\tan m x} \cdot \sec^2 m x}{4 \sqrt{\cos \sqrt{\tan m x}} \cdot \sqrt{\tan m x}} } \\ \)Please let me know in the comments if you find any errors in this solution.