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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.

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  • $\begingroup$ 0.5 in answer to b. should be 1/3. $\endgroup$ Mar 10, 2013 at 13:06
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    $\begingroup$ +1 for showing what you have done and checking the results with wolfram $\endgroup$ Mar 10, 2013 at 13:09
  • $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Mar 10, 2013 at 13:09
  • $\begingroup$ BTW, +1 for showing self effort and some real work. $\endgroup$
    – DonAntonio
    Mar 10, 2013 at 13:12
  • $\begingroup$ The Maple DiffTutor is your friend which produces these step by step with explanations. $\endgroup$
    – user64494
    Sep 14, 2013 at 7:42

4 Answers 4

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a. Multiply by $\sqrt{2+x}$ and something will simplify

b. Multiply by $(1-x^2)^{2/3}$.

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$$(a)\;\;\;\left(\frac{1-2x}{\sqrt{2+x}}\right)'=\frac{-2\sqrt{2+x}-\frac{1}{2\sqrt{2+x}}(1-2x)}{2+x}=\frac{-4(2+x)-(1-2x)}{2(2+x)^{3/2}}=\ldots$$

$${}$$

$$(b)\;\;\;\;\;\; \left(3x(1-x^2)^{1/3}\right)'=3(1-x^2)^{1/3}+3x(-2x)\frac{1}{3}(1-x^2)^{-2/3}=\ldots $$

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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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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

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