Find derivative (dy/dx) of function y = cos(x/(1+sqrt(x)))

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:

To find \( \frac{dy}{dx} \), we will apply the Chain Rule.

\( \displaystyle \text{Let } u = \frac{x}{1+\sqrt{x}}. \text{Then }y = \cos(u) \\ \displaystyle \frac{dy}{dx} = -\sin(u) \cdot \frac{du}{dx} = -\sin(\frac{x}{1+\sqrt{x}}) \cdot \frac{du}{dx} \\ \)

Now, compute \( \frac{du}{dx} \) using the quotient rule:

\( \begin{align*} \frac{du}{dx} &= \frac{(1+\sqrt{x}) \cdot \frac{d}{dx}(x) – x \cdot \frac{d}{dx}(1+\sqrt{x})}{(1+\sqrt{x})^2} \\ &= \frac{(1+\sqrt{x})(1) – x \left(\frac{1}{2\sqrt{x}}\right)}{(1+\sqrt{x})^2} \\ &= \frac{1 + \sqrt{x} – \frac{\sqrt{x}}{2}}{(1+\sqrt{x})^2} \\ &= \frac{1 + \frac{\sqrt{x}}{2}}{(1+\sqrt{x})^2} \\ &= \frac{2 + \sqrt{x}}{2(1+\sqrt{x})^2} \\ \end{align*} \)

Substitute \( \frac{du}{dx} \) back into \( \frac{dy}{dx} \):

\( \begin{align*} \frac{dy}{dx} &= -\sin\left(\frac{x}{1+\sqrt{x}}\right) \cdot \frac{2 + \sqrt{x}}{2(1+\sqrt{x})^2} \\ \end{align*} \)

Thus, the derivative of the function is:

\( \boxed{\frac{dy}{dx} = -\sin\left(\frac{x}{1+\sqrt{x}}\right) \cdot \frac{2 + \sqrt{x}}{2(1+\sqrt{x})^2}} \\ \)

Please let me know in the comments if you find any errors in this solution.