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