Question

Let \(u\) be a differentiable function of \(x\). Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\).

Short answer

The derivative of the absolute value function is \(u' \cdot u/|u|\), where |u| is expressed as \(\sqrt{u^2}\), and the chain rule of derivatives has been applied.

Step-by-step solution

Express Absolute Value in terms of Square Root

Recall that \(|u| = \sqrt{u^2}\). So we need to find the derivative \(d(\sqrt{u^2})/dx\).

Utilize Chain rule

Chain rule states that the derivative of \(f(g(x))\) where \(f\) and \(g\) are functions is \(f'(g(x)) \cdot g'(x)\). Here, apply the chain rule. The derivative of the square root function \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). So, the derivative of \(\sqrt{u^2}\) will be \(\frac{1}{2\sqrt{u^2}} \cdot (2u)\) (where \(2u\) comes from the derivative of \(u^2\)).

Simplify

Now simplifying this expression gives us \(u/|u|\), because \(2u\) and \(\frac{1}{2\sqrt{u^2}}\) cancel out the 2s, leaving \(u/|u|\). The derivative of \(u\) is represented as \(u'\), so the derivative of \(|u|\) or \(\sqrt{u^2}\) is \(u' \cdot u/|u|\), provided \(u \neq 0\) to avoid division by zero.

Conclusion

Hence, is has been proven that the derivative of the absolute value function is \(u' \cdot u/|u|\), where |u| is expressed as \(\sqrt{u^2}\), and the chain rule of derivatives has been applied.