Question

f(x)=4корень из x^3, f'(256)

Answer 1

$f(x) = 4 \sqrt{ {x}^{3} } = 4 {x}^{ \frac{3}{2} } \\ {f}^{)} (x) = 4 \times \frac{3}{2} \times {x}^{ \frac{3}{2} - 1 } = 6 {x}^{ \frac{1}{2} } = \\ 6 \sqrt{x} \\ {f}^{)} (256) = 6 \times \sqrt{256} = 6 \times 16 = 96$