2
$\begingroup$

I need to find first derivative of $x\sqrt{2-x^2}$. My approach

  1. Using product rule: $(2-x^2)^{1/2} + x\frac{\operatorname{d}(2-x^2)^{1/2}}{\operatorname{d}x}$

  2. Using chain rule: $(2-x^2)^{1/2} + x\left[\frac{1}{2}(2-x^2)^{-1/2} (-2x)\right]$

  3. Result: $(2-x^2)^{1/2} - x^2 (2-x^2)^{-1/2}$

    Is that correct?

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Yes, that is correct. It may be worth considering whether $$\frac{2(1-x^2)}{\sqrt{2-x^2}}$$ is a nicer form. $\endgroup$ – Daniel Fischer Nov 15 '13 at 20:18
  • $\begingroup$ Can you start using Latex for your math? You've asked enough questions on this site to do that, I believe. $\endgroup$ – Kaster Nov 15 '13 at 20:18
  • $\begingroup$ Is there any guide/manual to use Latex for this site? $\endgroup$ – J.Olufsen Nov 15 '13 at 20:20
  • 1
    $\begingroup$ Take a look here. But I'd just press edit on your old edited questions, so you can get the very basic commands as a starter. $\endgroup$ – Kaster Nov 15 '13 at 20:22
  • $\begingroup$ +1 for showing us your work! I cannot encourage that enough. Does my answer clear up the connections between the answers? $\endgroup$ – Cameron Buie Nov 15 '13 at 20:37
1
$\begingroup$

It is, indeed, correct. A way to see that the answers are the same is to note that $$\begin{align}(2-x^2)^\frac12-x^2(2-x^2)^{-\frac12} &= (2-x^2)^{1+-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= (2-x^2)(2-x^2)^{-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= \bigl((2-x^2)-x^2\bigr)(2-x^2)^{-\frac12}\\ &= \frac{2-2x^2}{(2-x^2)^\frac12}\\ &= \frac{2(1-x^2)}{\sqrt{2-x^2}}\\ &= -\frac{2(x^2-1)}{\sqrt{2-x^2}}.\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.