Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step:
Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$
Now, this is what I have done:
$$
(\frac{a+x}{a-x})^{\frac{3}{2}} \\
\implies \frac{\delta}{\delta y}\frac{f}{g} \\
\implies gf' = (a-x)^{\frac{3}{2}} \times \frac{3}{2} (a+x)^{\frac{1}{2}} \times 2 \\
\implies fg' \implies (a+x)^{\frac{3}{2}} \times \frac{3}{2} (a-x)^{\frac{1}{2}} \times 0 = 0 \\
\implies \frac{(a-x)^{\frac{3}{2}} \times 3 (a+x)^{\frac{1}{2}}}{(a-x)^3}\\
\implies \frac{(a-x)^{\frac{3}{2}} - 3\sqrt{a+x}}{(a-x)^3}
$$
But the answer is:
$$
\frac{3\times a (a+x)^{\frac{1}{3}}}{(a-x)^{\frac{5}{2}}}
$$
WolframAlpha shows:
$$
\frac{3a \sqrt{\frac{a+x}{a-x}}}{(a-x)^2}
$$
Another Answer (Somehow I got this):
$$
\frac{3 \sqrt{\frac{a+x}{a-x}}}{2(a-x)}
$$
==================
EDIT 1:
What about:
$$
y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\
y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x}\\
\implies \frac{3}{2}u^{\frac{1}{2}} \hspace{0.5cm} ; \hspace{0.5cm} \frac{(0+1)\times (a-x) - [ -1 (a+x) ]}{(a-x)^2} \\
\implies \frac{2a}{(a-x)^2} \\
\implies \frac{3}{2}\sqrt{\frac{2a}{(a-x)^2}} = \frac{3}{2} \times \frac{\sqrt{2a}}{a-x} 
$$

i think the right answer is this here
$$\frac{3}{2} \sqrt{\frac{a+x}{a-x}}
   \left(\frac{a+x}{(a-x)^2}+\frac{1}{a-x}\right)$$
