need help simplifying a differentiated quotient

Question

I am having trouble simplifying:

$$\frac{\left(\frac{1}{2\sqrt{x}} \cdot \cos(x)\right) - \left(\sqrt{x} \cdot -\sin(x)\right)}{\cos^2(x)}$$

The final simplification is:

$$\ \frac{cosx+2x\cdot\sin(x)}{2\sqrt{x}\cdot\cos^2(x)}$$

I am not sure how to get it to that and would like to know how it was simplified to that. THanks

Answer 1

One of the expressions in the question is not correct. The following is a simplification of the given expression. $$\begin{align*}\frac{\left(\dfrac{1}{2\sqrt{x}} \cdot \cos(x)\right) - \left(\sqrt{x} \cdot -\sin(x)\right)}{2\sqrt{x}\cdot\cos^2(x)} &= \frac{\left(\dfrac{1}{2\sqrt{x}} \cdot \cos(x)\right) + \left(\sqrt{x} \cdot \sin(x)\right)}{2\sqrt{x}\cdot\cos^2(x)} \\ &= \frac{\dfrac{\cos(x) + 2x\cdot\sin(x)}{2\sqrt{x}}}{2\sqrt{x}\cdot\cos^2(x)} \\ &= \dfrac{\cos(x)+2x\cdot\sin(x)}{4x\cdot\cos^2(x)}\end{align*}$$

Answer 2

Hint:

$$\frac{\frac{A}{B}- C}{D}=\frac{\frac{A-BC}{B}}{D}=\frac{A-BC}{B}\cdot\frac1D=\frac{A-BC}{BD}$$