<s>[INST] Find the derivative at (1,2) 
$$f(x) = x^2 \sqrt{5 - x^2}$$
Find the derivative at $(1, 2)$.
\begin{align}
\frac{d}{dx} \left[ x^2 \sqrt{5 - x^2} \right] & = \frac{d}{dx} \left[ x^2 (5 - x^2)^{1/2} \right] \\
& = x^2 \frac12 (5 - x^2)^{-1/2}(-2x) + (5 - x^2)^{1/2}(2x)
\end{align}

The equation is formed using the product rule and the chain rule. The author explained that $(-2x)$ was inserted as "multiply as the inside derivative", and $(2x)$ as the derivative of the first term.
I assume the following:
$(-2x)$ is the derivative of the inside derivative which is $(5-x^2)$. 
$(2x)$ is the derivative of $x^2$.
My question is, why isn't $(-2x)$ multiplied for the 2nd part of the equation (after the $+$ sign)?
 [/INST] The derivative actually reads $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x)=x^2$ and $v(x)=\sqrt{5-x^2}$. 
For $u(x)$, we have $u'(x)=2x$. 
For $v(x)$, we can rewrite $v(x)=f(g(x))$ where $f(t)=\sqrt{t}$ and $g(x)=5-x^2$, so that $v'(x)=g'(x)f'(g(x))=-\frac{2x}{2\sqrt{5-x^2}}$.
Hence $f'(x)=2x\sqrt{5-x^2}-(x^2)(-\frac{2x}{2\sqrt{5-x^2}})=2x\sqrt{5-x^2}+x^2\frac{2x}{2\sqrt{5-x^2}}$, which is the equation shown.
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