Question

Absolute Value Functions Let \(u\) be a differentiable function of \(x .\) (a) Show that \(\frac{d}{d x}|u|=u^{\prime} \frac{u}{|u|}\) (b) Use part (a) to find the derivatives of \(f(x)=\left|x^{2}-9\right|\) and \(g(x)=|x| \sin x .\)

Short answer

The derivatives of the given functions are \(f'(x) = \frac{2x}{|x|}\) for \(f(x) = |x^2 - 9|\) and \(g'(x) = x/|x| * sin(x) + |x| * cos(x)\) for \(g(x) = |x| sin(x)\).

Step-by-step solution

Derive general formula for Absolute Value Function Derivative

The absolute value of \(u\) is defined as \(|u| = \sqrt{u^2}\). Differentiate this with respect to \(u\), to verify the formula given. In terms of chain rule, the differentiation will be \(\frac{d}{du}(u^2)^{1/2} = \frac{1}{2}(u^2)^{-1/2} * 2u * u' = \frac{u}{|u|} * u'\). This verifies our formula \(\frac{d|u|}{dx} = \frac{u}{|u|} u'\)

Apply formula to function \(f(x) = |x^2 - 9|\)

Differentiate function \(f(x) = |x^2 - 9|\) with respect to \(x\), using the formula derived in step 1. Apply chain rule and the result will be \(\frac{d|u|}{dx} = \frac{u'}{|u|}*u = \frac{(2x)'}{|2x|}*(2x) = \frac{2*x*(2)}{|2x|} = \frac{4x}{2|x|} = \frac{2x}{|x|},\) where \(u = x^2 - 9\).

Apply formula to function \(g(x) = |x| sin(x)\)

To differentiate \(g(x) = |x| sin(x)\), apply product rule first \(h'(x) = f'(x)g(x) + f(x)g'(x)\), let \(f(x) = |x|\) and \(g(x) = sin(x)\). So, \(g'(x) = f'(x) * sin(x) + |x| * cos(x)\). For \(f'(x) = |x|',\) apply the formula derived in step 1 to get \(\frac{x}{|x|}\). Therefore, \(g'(x) = x/|x| * sin(x) + |x| * cos(x)\).