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?

We are given   h(x) = |8x| · cos(4x). We want to differentiate h(x) using the product rule and the result   d/dx |u| = u′ · (u/|u|),  (for u ≠ 0). Step 1. Write h(x) as a product of two functions:   u(x) = |8x|  and  v(x) = cos(4x). Step 2. Differentiate u(x). Here, the inner function is 8x (which is differentiable) and we apply the given result by letting   w(x) = 8x  so that  |8x| = |w(x)|. Then, using the given rule,   d/dx |w(x)| = w′(x) · (w(x)/|w(x)|). Since w(x) = 8x, we have w′(x) = 8 and thus, for x ≠ 0,   d/dx |8x| = 8 · (8x/|8x|) = 8 · (x/|x|)  [Note: since |8x| = 8|x|, the factor 8 cancels out in the fraction]. So,   u′(x) = 8·(x/|x|)  for x ≠ 0. Step 3. Differentiate v(x). We have   v(x) = cos(4x) and by the chain rule,   v′(x) = -sin(4x)·4 = -4 sin(4x). Step 4. Now, use the product rule:   h′(x) = u′(x)v(x) + u(x)v′(x). Substitute the expressions we found:   h′(x) = [8·(x/|x|)] · cos(4x) + |8x| · [ -4 sin(4x)]. Since |8x| = 8|x|, we have:   h′(x) = 8 (x/|x|) cos(4x) - 4·8|x| sin(4x)        = 8 (x/|x|) cos(4x) - 32 |x| sin(4x). Thus, for x ≠ 0, the derivative is:   h′(x) = 8 (x/|x|) cos(4x) - 32 |x| sin(4x). Answer in boxed form:   h′(x) = [8 (x/|x|) cos(4x) - 32 |x| sin(4x)]  (for x ≠ 0).