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. \[ h(x)=|8 x| \cos (4 x) \] \( h^{\prime}(x)=\square \)

We are given   h(x) = |8x| cos(4x) and the fact that if u = u(x) is differentiable with u(x) ≠ 0, then   d/dx [|u|] = u′(x) · (u(x)/|u(x)|). Step 1. Write h(x) as a product:   h(x) = |8x| · cos(4x). Step 2. Differentiate using the product rule:   h′(x) = (d/dx |8x|) · cos(4x) + |8x| · (d/dx cos(4x)). Step 3. Compute each derivative. (a) Differentiate |8x|:  Let u(x) = 8x so that u′(x) = 8.  Then, for x ≠ 0,   d/dx |8x| = 8 · (8x / |8x|).  Notice that   8x/|8x| = (8x)/(8|x|) = x/|x|.  Thus,   d/dx |8x| = 8(x/|x|). (b) Differentiate cos(4x):  d/dx cos(4x) = -sin(4x) · (4) = -4 sin(4x). Step 4. Substitute these results back into the product rule:  h′(x) = [8 (x/|x|)] cos(4x) + |8x| · [–4 sin(4x)].  That is,   h′(x) = 8 cos(4x) · (x/|x|) − 4|8x| sin(4x). Step 5. (Optional) Simplify by noting that |8x| = 8|x|:  h′(x) = 8 cos(4x) · (x/|x|) − 4(8|x|) sin(4x)      = 8 cos(4x) · (x/|x|) − 32 |x| sin(4x). Thus, the derivative of h(x) is:   h′(x) = 8 cos(4x) · (x/|x|) − 32|x| sin(4x)   for x ≠ 0. A piecewise answer (since x/|x| equals 1 for x > 0 and –1 for x < 0) is:  For x > 0:   h(x) = 8x cos(4x) ⇒ h′(x) = 8 cos(4x) − 32x sin(4x).  For x < 0:   h(x) = −8x cos(4x) ⇒ h′(x) = −8 cos(4x) + 32x sin(4x). Either form is acceptable.