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
 A: There is one trick which is very useful when you face products, quotients, powr,.. : logarithmic differentiation.
$$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}\implies \log(f(x))=\frac 12 \log(3-x^2)-\log(3+x)$$
$$\frac{f'(x)}{f(x)}=\frac 12\frac{-2x}{3-x^2}-\frac 1{3+x}=-\frac{3 (x+1)}{(x+3) \left(3-x^2\right)}$$
Now
$$f'(x)=f(x) \times \frac{f'(x)}{f(x)}$$ Just simplify.
A: Your derivative of $\sqrt{3 - x^2}$ is incorrect.
$$\begin{align}\dfrac{\mathrm d}{\mathrm dx}\dfrac{\sqrt{3 - x^2}}{3 + x^2} &= \dfrac{\frac{\mathrm d}{\mathrm dx}\sqrt{3 - x^2}\cdot(3 + x) - \sqrt{3 - x^2}\cdot\frac{\mathrm d}{\mathrm dx}(3 + x)}{(3 + x)^2} \\ &= \dfrac{\frac{-2x}{2\sqrt{3 - x^2}}(3 + x) - \sqrt{3 - x^2}}{(3 + x)^2} \\ &= \dfrac{-x(3 + x) - (3 - x^2)}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{x^2 - x^2 - 3x - 3}{(3 + x)^2\sqrt{3 - x^2}} = -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}$$
A: your first line is not entirely clear 
\begin{eqnarray*}
f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}} \color{red}{ \left( \frac{d}{dx}(3-x^{\color{blue}2}) \right)}(3+x)-\sqrt{3-x^2}}{(3+x)^2}
\end{eqnarray*}
But everything is right after this and you just need to do a common denominator at the end & you will get the Wolfie answer ... look again, there is a minus sign!
A: Yet another method in case you're interested: first square both sides
$$y^2=\frac{3-x^2}{(3+x)^2}$$
Then differentiate both sides:
$$2y\frac{dy}{dx}=\frac{(3+x)^2(-2x)-(3-x^2)2(3+x)}{(3+x)^4}$$
$$=-\frac{6(x+1)}{(3+x)^3}$$
Then divide both sides by $2y$ to get the result.
A: $f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$ is false ! 
Correct is:
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(-x^2)-\sqrt{3-x^2}}{(3+x)^2}$.
Now proceed !
A: You just need to take the LCM of the denominators.
$$\begin{align}\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2} &= \frac{-x(3 + x) - \sqrt{3 - x^2}\sqrt{3 - x^2}}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{-3x - x^2 - 3 + x^2}{(3 + x)^2\sqrt{3 - x^2}} \\ &= -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}$$
A: Wow. There are many good solutions here. 
@James Warthington, I'd like to add another approach but using a more methodical application of the Chain Rule. 
$y=(3+x)^{-1}(3-x^2)^ \frac{1}{2}$
First variable substitution, let $\mu=3+x$, and note(for later) that $\frac{d}{dx}\mu=1$
Second variable substitution, let $\omega=3-x^2$, and note(for later) that $\frac{d}{dx}\omega=-2x$
Our equation is now $y=\mu^{-1}\omega^ \frac{1}{2}$. Now let's apply the Differential Operator to both sides...
$\frac{d}{dx}y=\frac{d}{dx}[\mu^{-1}\omega^ \frac{1}{2}]$. $\space\space\space$ Apply the multiplicative operation of derivatives...
$y'=\mu^{-1}\frac{d}{dx}\omega^ \frac{1}{2}+\omega^ \frac{1}{2}\frac{d}{dx}\mu^{-1}$
You cannot $\frac{d}{dx}\mu$ or $\frac{d}{dx}\omega$  directly. You must apply the Chain Rule!  This rule essentially says you can change the $x$ to $\mu$ but only if you multiply by $\frac{d\mu}{dx}$.  Same thing applies to $\omega$.
$y'=\mu^{-1}\frac{d}{d\omega}\omega^ \frac{1}{2}\frac{d\omega}{dx}+\omega^ \frac{1}{2}\frac{d}{d\mu}\mu^{-1}\frac{d\mu}{dx}$
Now, this is very straight-forward stuff.  
$\frac{d}{d\omega}\omega^ \frac{1}{2}$ is $\frac{1}{2\sqrt\omega}$ $\space\space\space$...and...$\space\space\space$ $\frac{d}{d\mu}\mu^{-1}$ is $\frac{-1}{\mu^{2}}$.  $\space\space\space$Now let's substitute everything...
$y'=\left(\mu^{-1}\right)\left[\frac{1}{2\sqrt\omega}\right]\left(-2x\right)+ \omega^ \frac{1}{2} \left[ \frac{-1}{\mu^{2}} \right] \left(1\right)$. $\space\space\space\space\space$Simplifying...
$y'=\frac{-2x}{2\mu\sqrt\omega}$ + $\frac{-1\sqrt\omega}{\mu^2}$ $\space\space\space$...and factor out $\frac{-1}{\mu}$ ...$\space\space\space$ $y'=\left(\frac{-1}{\mu}\right)\left[\frac{x}{\sqrt\omega}+\frac{\sqrt\omega}{\mu}\right]$ 
Get a common denominator in the square brackets and multiply through...
$y'=-\frac{x\mu+\omega}{\mu^2\sqrt\omega}$. $\space\space\space\space\space$Now re-substitute variables...
$y'=-\frac{x(3+x)+\left(3-x^2\right)}{(3+x)^2\sqrt{3-x^2}}$
Expand and factor the numerator...
$y'=-\frac{3x+x^2+3-x^2}{(3+x)^2\sqrt{3-x^2}}$
$y'=-\frac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
The technique used by @Claude Leibovici, logarithmic differentiation, is very interesting. I've never come across this approach and will certainly study it and add it to my bag of tricks!
