Question

f(x)=4корень из x^3, f'(256)

Answer 1

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