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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.
a. Multiply by $\sqrt{2+x}$ and something will simplify
b. Multiply by $(1-x^2)^{2/3}$.
$$(a)\;\;\;\left(\frac{1-2x}{\sqrt{2+x}}\right)'=\frac{-2\sqrt{2+x}-\frac{1}{2\sqrt{2+x}}(1-2x)}{2+x}=\frac{-4(2+x)-(1-2x)}{2(2+x)^{3/2}}=\ldots$$
$${}$$
$$(b)\;\;\;\;\;\; \left(3x(1-x^2)^{1/3}\right)'=3(1-x^2)^{1/3}+3x(-2x)\frac{1}{3}(1-x^2)^{-2/3}=\ldots $$
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.) $$(3x(1-x^2)^{1/3})'=3(1-x^2)^{1/3}-2x^2(1-x^2)^{-2/3}$$
There's actually a trick for simplifying part (a). To simplify, pull out the lowest power of (2 + x) that appears in each term. This would be $\ (2 + x)^{-1/2}$. What's left over is almost certainly something that can be combined or further simplified and is neater. This video has more help on using the quotient rule if you want more help: http://www.youtube.com/watch?v=jIX0VvwfEko
