<s>[INST] finding the derivative using quotient rule and product rule find dy/dx;
a) $\frac{1-2x}{\sqrt{2+x}}$
b.) $3x(1-x^2)^{1/3}$
My attempt at a)
use the quotient rule:
so $dy/dx = -2 \sqrt{2+x}+ (1-2x)0.5(2+x)^{-1/2}$
but then I get stuck there, cannot simplify it, wolfram gives a nice simplified answer but not sure how to get it.
b.) Product rule:
$dy/dx= 3(1-x^2)^{0.5} + 3 \times 1/3 \times (1-x^2)^{-2/3}$ and again i can't seem to simplify that either to a nice wolfram answer. 
 [/INST] a) $$\left(\frac{1-2x}{\sqrt{2+x}}\right)'=\frac{-2\sqrt{2+x}-(1-2x)\frac{1}{2\sqrt{2+x}}}{2+x}=\frac{-4(2+x)-(1-2x)}{2\sqrt{2+x}(2+x)}$$
$$=\frac{-9-2x}{2\sqrt{2+x}(2+x)}=-\frac{9+2x}{2(2+x)^{3/2}}$$
b.) $$(3x(1-x^2)^{1/3})'=3(1-x^2)^{1/3}-2x^2(1-x^2)^{-2/3}$$
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