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.

• 0.5 in answer to b. should be 1/3. – Emanuele Paolini Mar 10 '13 at 13:06
• +1 for showing what you have done and checking the results with wolfram – Dominic Michaelis Mar 10 '13 at 13:09
• your attempt at (a) is missing divison by the denominator squared...and in (b) the first summand has an exponent $\,1/3\,$ , not $\,0.5\,$ , and the second summand lacks the inner derivative... – DonAntonio Mar 10 '13 at 13:09
• BTW, +1 for showing self effort and some real work. – DonAntonio Mar 10 '13 at 13:12
• The Maple DiffTutor is your friend which produces these step by step with explanations. – user64494 Sep 14 '13 at 7:42

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