Derivative of absolute value of a function

Question

What would be the time derivative of $|y(x(t))|^\alpha$ where $\alpha \geq 1$?? Would a $\operatorname{sign}(y(x(t)))$ show up in the derivative?? I am confused about this.

Answer 1

HINT:

A nice trick is that the real modulus function can be given by $x \mapsto \sqrt{x^2}$.

For example, if $y=|x|$, then $y \equiv \left(x^2\right)^{1/2}$.

Applying the chain rule gives:

\begin{eqnarray*} \frac{\mathrm dy}{\mathrm dx} &=& \frac{1}{2}\left(x^2\right)^{-1/2}\times 2x \\ \\ &\equiv& \frac{x}{\sqrt{x^2}} \\ \\ &\equiv& \frac{x}{|x|} \end{eqnarray*}

Similar tricks will allow you to differentiate $\mathrm f(|x|)$, $|\mathrm f(x)|$ and even $|\mathrm f(|x|)|$.

The sign function is exactly $\displaystyle{\frac{x}{|x|}}$.