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Ответ:

Пошаговое объяснение:

1.

f'(x)=(6x²-3x+1)'=12x-3=3*(4x-1).

f'(x)=(x³*sinx)'=2x*sinx+x³*cosx=x*(2*sinx+x²*cosx).

2.

f'(x)=(x²/(1+x²))'=(2x*(1+x²)-x²*2x)/(1+x²)²=(2x*(1+x²-x²)/(1+x²)²=2x/(1+x²)².

3.

y=x²-7x+12       x₀=3

yk=y(x₀)+y'(x₀)*(x-x₀)

y(3)=3²-7*3+12=9-21+12=0

y'(3)=(y²-7x+12)'=2y-7=2*3-7=6-7=-1.   ⇒

yk=0-1*(x-3)

yk=3-x.

4.

x(t)=(t²/2)+6x+23

v=x'(t)=((t²/2)+6t+23)'=t+6

t+6=19

t=13 (c).