Question

найдите вторую производную f(x)=sin²x, f(x)=cos2x
f(x)=корень из x, f(x)=x²- 2корень из x
f(x)= xsinx, f(x)=xcos3x, f=3x²-cos(x²+1)
f(x)=sin²2x f(x)=x²sin2x

Answer 1

Ответ:

1

$f(x) = \sin {}^{2} (x)$

$f'(x) = 2 \sin(x) \times (\sin(x) ) '= 2 \sin(x) \cos(x) = \sin(2x)$

2

$f(x) = \cos(2x)$

$f'(x) = - \sin(2x) \times (2x) '= - 2 \sin(2x)$

3

$f(x) = \sqrt{x} = {x}^{ \frac{1}{2} }$

$f'(x) = \frac{1}{2} {x}^{ - \frac{1}{2} } = \frac{1}{2 \sqrt{x} }$

4

$f(x) = {x}^{2} - 2 \sqrt{x}$

$f'(x) = 2x - 2 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = 2x - \frac{1}{ \sqrt{x} }$

5

$f(x) = x \sin(x)$

$f'(x) = x' \sin(x) + ( \sin(x) ) '\times x = \\ = \sin(x) + x \cos(x)$

6

$f(x) = x \cos(3x)$

$f'(x) = x' \cos(3x) + ( \cos(3x) ) '\times x = \\ = \cos(3x) - 3x\sin(3x)$

7

$f(x) = 3 {x}^{2} - \cos( {x}^{2} + 1 )$

$f'(x) = 6 + \sin( {x}^{2} + 1) \times ( {x}^{2} + 1) '= \\ = 6 + 2x \sin( {x}^{2} + 1)$

8

$f(x) = \sin {}^{2} (2x)$

$f'(x) = 2 \sin(2x) \times ( \sin(2x) )' \times (2x) = \\ = 2 \sin(2x) \times \cos(2x) \times 2 = 2 \sin(4x)$

9

$f(x) = {x}^{2} \sin(2x)$

$f'(x) = 2x \sin(2x) + \cos(2x) \times 2 \times {x}^{2} = \\ = 2x \sin(3x) + 2 {x}^{2} \cos( 3x)$

Answer 2

1) f'(x)=(sin²x)'=2*sinx*cosx=sin2x; f''(x)=(sin2x)'=2cos2x;

2)f'(x)=(cos2x)'=-2*sin2x; f''(x)=(-2sin2x)'=-4cos2x

3) f'(x)=(√x)'=1/(2√x)=(x⁻¹/²)/2; f''(x)=((x⁻¹/²)/2)'=-(1/4)*x⁻³/²=-1/(4x√x)

4)  f'(x)=(x²-2√x)'=2x-(2/(2√x))=(2x-x⁻¹/²); f''(x)=(2x-(x⁻¹/²))'=2-(-1/2)*x⁻³/²=

2+1/(2x√x);

5) f'(x)= (xsinx)'=sinx+x*cosx; f''(x)=сosx+cosx-x*sinx=2cosx-x*sinx;

6) f'(x)=(xcos3x)'=cos3x-3x*sin3x; f''(x)=(cos3x-3x*sin3x)'=-3sin3x-3sin3x-9x*сos3x=-6sin3x-9x*сos3x;

7)  f'(x)=(3x²-cos(x²+1))'=6x+sin(x²+1)*(2x)=2x*(3+sin(x²+1)); f''(x)=

(2x*(3+sin(x²+1)))'=

2*(3+sin(x²+1))+2x*(2x*cos(x²+1))=6+2sin(x²+1)+4x²*cos(x²+1);

8) f'(x)=(sin²2x)'=2*sin2x*(cos2x)*2=2sin4x; f''(x)=(2sin4x)'=8*cos4x;

9)  f'(x)=(x²sin2x)'=2x*sin2x+2x²*cos2x;

f''(x)=(2x*sin2x+2x²*cos2x)'=2*sin2x+4x*cos2x+4x*cos2x-4x²*sin2x=

2*sin2x+8x*cos2x-4x²*sin2x