Question

Вычислите значение производной функции ƒ(x) =5x^2* корень x - 64/ корень x^3

Answer 1

$f(x) = 5 {x}^{2} \sqrt{x } - \frac{64}{ \sqrt{ {x}^{3} } } = 5 {x}^{2 + \frac{1}{2} } - 64 {x}^{ - \frac{3}{2} } = 5 {x}^{ \frac{5}{2} } - 64 {x}^{ - \frac{3}{2} }$
f'(x) =
$5 \times \frac{5}{2} \times {x}^{ \frac{5}{2} - 1 } - 64 \times ( - \frac{3}{2} ) \times {x}^{ - \frac{ 3}{2} - 1 } = \frac{25}{2} {x}^{ \frac{3}{2} } + 96 {x}^{ - \frac{5}{2} }$
f'(4) =
$\frac{25}{2} \times {4}^{ \frac{3}{2} } + 9 6 \times {4}^{ - \frac{5}{2} } = \frac{25}{2} \times \sqrt{ {4}^{3} } + \frac{96}{ \sqrt{ {4}^{5} } } = \frac{25}{2} \times \sqrt{64} + \frac{96}{ \sqrt{1024} } = \frac{25}{2} \times 8 + \frac{96}{32} = 25 \times 4 + 3 = 100 + 3 =103$
Ответ: 103

Answer 2

F(x)=5x²√x-64/√x³
f'(x)=5*(x^5/2)'-64*(x^(-3/2)'=

5*5/2*x^(5/2-1)-64*(-3/2)*x^(-3/2-1)=
25/2*x^(3/2)+32*3*x^(-5/2)=

25/2*x^(3/2)+96/x^5/2=
(12,5x^(3/2+5/2)+96)/x²√x=
(12,5x⁴+96)/x²√x
f'(4)=(12,5*256+96)/16*2=
103