The given function is \( \displaystyle y = \sin\left(\sqrt{\sin x + \cos x}\right) \) and we need to find the derivative of this function with respect to \( x \).
Solution:
Let \(\displaystyle u = \sqrt{\sin x + \cos x} \). Then, \( y = \sin u \).
Therefore, using the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{d}{du}(\sin u) \cdot \frac{du}{dx} = \cos u \cdot \frac{du}{dx} \dots (1) \\ \)Let \( v = \sin x + \cos x \), then \( u = \sqrt{v} = v^{1/2} \).
Therefore, using the chain rule:
\( \displaystyle \frac{du}{dx} = \frac{d}{dv}v^{1/2} \cdot \frac{dv}{dx} = \frac{1}{2}v^{-1/2} \cdot \frac{dv}{dx} \\ \)Now differentiate \( v = \sin x + \cos x \):
\( \displaystyle \frac{dv}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) = \cos x – \sin x \\ \)Substitute the values back in (1) to find the derivative:
\( \begin{aligned} \frac{dy}{dx} &= \cos u \cdot \frac{du}{dx} \\ &= \cos u \cdot \frac{1}{2}v^{-1/2} \cdot \frac{dv}{dx} \\ &= \cos u \cdot \frac{1}{2}v^{-1/2} \cdot (\cos x – \sin x) \\ &= \cos\left(\sqrt{\sin x + \cos x}\right) \cdot \frac{1}{2}(\sin x + \cos x)^{-1/2} \cdot (\cos x – \sin x) \\ &= \cos\left(\sqrt{\sin x + \cos x}\right) \cdot \left(\frac{\cos x – \sin x}{2\sqrt{\sin x + \cos x}}\right) \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{(\cos x – \sin x) \cdot \cos\left(\sqrt{\sin x + \cos x}\right)}{2\sqrt{\sin x + \cos x}}} \\ \)Please let me know in the comments if you find any errors in this solution.