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 \(\displaystyle u = \frac{x}{1+\sqrt{x}} \), then \( y = \cos(u) \).
First, find \(\displaystyle \frac{du}{dx} \) using the quotient rule:
\( \displaystyle \frac{du}{dx} = \frac{(1+\sqrt{x})\cdot \frac{d}{dx}(x)\, – x \cdot \frac{d}{dx}(1+\sqrt{x})}{(1+\sqrt{x})^2} \\ \displaystyle = \frac{(1+\sqrt{x})\cdot 1\, – x \cdot \left(\frac{1}{2\sqrt{x}}\right)}{(1+\sqrt{x})^2} \\ \displaystyle = \frac{1+\sqrt{x}\, – \frac{\sqrt{x}}{2}}{(1+\sqrt{x})^2} \\ \displaystyle = \frac{1 + \frac{\sqrt{x}}{2}}{(1+\sqrt{x})^2} \\ \displaystyle = \frac{2 + \sqrt{x}}{2(1 + \sqrt{x})^{2}} \\ \)Now, apply the chain rule to find \(\displaystyle \frac{dy}{dx} \):
\( \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \\ \displaystyle = -\sin(u) \cdot \frac{2 + \sqrt{x}}{2(1 + \sqrt{x})^{2}} \\ \displaystyle = -\sin\left(\frac{x}{1+\sqrt{x}}\right) \cdot \frac{2 + \sqrt{x}}{2(1 + \sqrt{x})^{2}} \\ \)Thus, the derivative of the function is:
\( \displaystyle \boxed{\frac{dy}{dx} = -\sin\left(\frac{x}{1+\sqrt{x}}\right) \cdot \frac{2 + \sqrt{x}}{2(1 + \sqrt{x})^{2}}} \\ \)Please let me know in the comments if you find any errors in this solution.