Finding derivative using product and chain rule I need to find first derivative of $x\sqrt{2-x^2}$. My approach

*

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


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


*Result: $(2-x^2)^{1/2} - x^2 (2-x^2)^{-1/2}$
Is that correct?


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