Question

1. Найти значение производной в точке х0
А) f(x)=1/6x^6-3x^2+2, x0=2
Б) f(x)=x^2+1/x-1, x0=-1
B) f(x)=(x^3-4)*(3x^2+1), x0=2
Г) f(x)=5x*cosx+2, x0=pi/2

Answer 1

A) x^5-6x
f(2)= 2^5-6*2
32-12=20
б) 2x-1/
f(-1)=-3

Answer 2

F'(x)=1/6 *6x^5 -3*2x = x^5-6x

f'(2)=2^5-6*2=32-12=20


f'(x)=2x-1/x^2

f'(-1)=2*(-1) -1/(-1)^2 = -2-1=-3


f'(x)=3x^2 *(3x^2+1) + (x^3-4)*6x

f'(2)=3*2^2 *(3*2^2+1)+(2^3-4)*6*2= 12*13+48=156+48=204


f'(x)=5*cosx -5x*sinx

f'(p/2) = 5 cos p/2 - 5p/2 * sin p/2 = 5*0 - 5p/2*1= -5p/2.