# Finding derivative using product and chain rule

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?

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

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}