2. find the derivative of ( y = \frac{sin sqrt{x}}{1+cos sqrt{x}} )

Explanation:

Step1: Use the quotient rule

The quotient rule states that if (y=\frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Let (u = \sin\sqrt{x}) and (v=1 + \cos\sqrt{x}). First, find (u^\prime): Using the chain rule, if (u=\sin t) and (t = \sqrt{x}=x^{\frac{1}{2}}), then (u^\prime=\cos t\cdot t^\prime). Since (t^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}), so (u^\prime=\frac{\cos\sqrt{x}}{2\sqrt{x}}). Next, find (v^\prime): If (v = 1+\cos t) and (t=\sqrt{x}), then (v^\prime=-\sin t\cdot t^\prime). Since (t^\prime=\frac{1}{2\sqrt{x}}), so (v^\prime=-\frac{\sin\sqrt{x}}{2\sqrt{x}}).

Step2: Apply the quotient rule

[ \begin{align*} y^\prime&=\frac{\frac{\cos\sqrt{x}}{2\sqrt{x}}(1 + \cos\sqrt{x})-\sin\sqrt{x}\left(-\frac{\sin\sqrt{x}}{2\sqrt{x}}\right)}{(1 + \cos\sqrt{x})^{2}}\ &=\frac{\frac{\cos\sqrt{x}+\cos^{2}\sqrt{x}+\sin^{2}\sqrt{x}}{2\sqrt{x}}}{(1 + \cos\sqrt{x})^{2}} \end{align*} ] Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), we have: [ \begin{align*} y^\prime&=\frac{\frac{\cos\sqrt{x}+ 1}{2\sqrt{x}}}{(1 + \cos\sqrt{x})^{2}}\ &=\frac{1}{2\sqrt{x}(1 + \cos\sqrt{x})} \end{align*} ]

Answer:

(y^\prime=\frac{1}{2\sqrt{x}(1 + \cos\sqrt{x})})