Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29

Your derivative of $\sqrt{3 - x^2}$ is incorrect.
$$\begin{align}\dfrac{\mathrm d}{\mathrm dx}\dfrac{\sqrt{3 - x^2}}{3 + x^2} &= \dfrac{\frac{\mathrm d}{\mathrm dx}\sqrt{3 - x^2}\cdot(3 + x) - \sqrt{3 - x^2}\cdot\frac{\mathrm d}{\mathrm dx}(3 + x)}{(3 + x)^2} \\ &= \dfrac{\frac{-2x}{2\sqrt{3 - x^2}}(3 + x) - \sqrt{3 - x^2}}{(3 + x)^2} \\ &= \dfrac{-x(3 + x) - (3 - x^2)}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{x^2 - x^2 - 3x - 3}{(3 + x)^2\sqrt{3 - x^2}} = -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}$$
