Let \( u \) be a differentiable function of \( x \). Assume the following result is proved. \[ \frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, u \neq 0 \text {. } \] Use this result to find the derivative of the function. \( h(x)=|8 x| \cos (4 x) \) \( h^{\prime}(x)=\square \) ReadIt Whitchit na nel Help?

To find the derivative of the function \( h(x) = |8x| \cos(4x) \), we will apply the product rule along with the given result for the derivative of the absolute value function. First, let \( u = 8x \). Then, by the product rule: \[ h'(x) = \frac{d}{dx}[|u|] \cos(4x) + |u| \frac{d}{dx}[\cos(4x)] \] From the result provided, we know: \[ \frac{d}{dx}[|u|] = u' \frac{u}{|u|} \quad \text{for } u \neq 0. \] Here, \( u' = 8 \) and \( u = 8x \), so: \[ \frac{d}{dx}[|8x|] = 8 \frac{8x}{|8x|} = 8 \cdot \text{sgn}(8x), \] where \( \text{sgn}(x) \) is the sign function. Next, the derivative of \( \cos(4x) \) is: \[ \frac{d}{dx}[\cos(4x)] = -4 \sin(4x). \] Now we can combine all components: \[ h'(x) = 8 \cdot \text{sgn}(8x) \cos(4x) + |8x| (-4 \sin(4x)). \] Thus, the final expression for the derivative \( h'(x) \) is: \[ h'(x) = 8 \cdot \text{sgn}(8x) \cos(4x) - 4|8x| \sin(4x). \]