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I need help solving this derivative using the chain rule. I have tried setting $u = -x^2$
2 Answers
$\begingroup$\begin{align*} \frac{d}{dx}\ |u| &= \frac{d}{dx} \ \sqrt{u^2} \\&= \frac{1}{2} (u^2)^\frac{-1}{2} \ \bigg(2u\frac{d}{dx} u \bigg) \\&= \frac{1}{|u|} (uu') \end{align*}
$\endgroup$ 3 $\begingroup$$\frac{d|u(x)|}{dx}=\frac{d|u|}{du}\frac{du}{dx}$ by the chain rule.
So, we need only examine the derivative
$$\frac{d|u|}{du}$$
Note that for $u>0$ the derivative is $+1$ while for $u<0$, the derivative is $-1$.
The derivative at $0$ is undefined since the left-sided and right-sided derivatives are not equal.
So, for $u\ne 0$, we can write
$$\frac{d|u(x)|}{dx}=\frac{du(x)}{dx}\times \frac{u}{|u|}$$
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