Question

42.2. Найдите значение производной функции в точке х0:
1) f(x) = 2cosx + sinx, x0 = п/3;
2) f(x) = 2х – cosx + 3sinx, x0 =п/4;
3) f(x) = х^2 – 2cosx + sinx, x0 =3п/4;
4) f(x) = 2/х + 2cosx + 4sinx, x0=2п/3.​

Answer 1

1) f'(x)= -2sin(x)+cos(x)= -2sin(p/3)+ cos(p/3)= -√3+0,5

2) f'(x)= 2+sin(x)+3cos(x)= 2+sin(p/4)+3cos(p/4)= 2+ √2/2 + √2/2= 2+√2

3) f'(x)= 2x+2sin(x)+cos(x)= 3π/2+2sin(3p/4)+cos(3p/4)= 3p/2 + √2 -√2/2= 3p/2+√2/2

4) f'(x)= -2/x² -2sin(x)+4cos(x)= -2/(4p²/9) -2sin(2p/3)+ 4cos(2p/3)= -9/2p² -√3-2

Answer 2

Ответ:

Объяснение:

1) f'=-2sinx+cosx,  f(p/3)=-2sinp/3+cosp/3=-2*V3/2+1/2=-V3+1/2

2)f'=2+sinx+3cosx,  f'(p/4)=2+sinp/4+3cosp/4=2+ V2/2+3*V2/2=2+ 4*V2/2=

2+ 2V2

3)f'=2x+2sinx+cosx,  f'(3p/4)=2*3p/4+2*sin3p/4+cos3p/4=3p/2+2*V2/2-V2/2=

3p/2+V2/2=3p+V2 /2

4)f'=-2/x^2-2sinx+4cosx,  f'(2p/3)=-2:(4p^2/9)-2sin2p/3+4cos2p/3=-9/2p^2-2*

V3/2+4*(-1/2)=-9/2p^2-V3-2