Решите 5 любых, Срочно!!

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Решите 5 любых, Срочно!!


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Алгебра | 26 просмотров
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Правильный ответ

1.2
f`(x)=(2+2x²-4x²)/(1+x²)²=(2-2x²)/(1+x²)²
f`(-3)=(2-18)/(1+9)²=-16/100=-0,16
1.3
f`(x)=(x-1)`*√(x²+1)+(x-1)*(√(x²+1))`=1*√(x²+1)+(x-1)*2x/(2√(x²+1))=
=√(x²+1)+x(x-1)/√(x²+1)=(x²+1+x²-x)/√(x²+1)=(2x²-x+1)/√(x²+1)
f`(0)=(0-0+1)/√(0+1)=1/1=1
1.4
y`=1/3*3cos²x*(-sinx)+sinx=sinx*(-cos²x+1)
y`(π/6)=sinπ/6*(-cos²π/6+1)=1/2*(-3/4+1)=1/2*1/4=1/8
1.5
f`(x)=4(x+1)/(x-1)*(x+1-x+1)/(x+1)²=4(x+1)/(x-1)*2/(x+1)²=8/[(x-1)(x+1)]=8/(x²-1)
f`(-3)=8/(9-1)=8/8=1
1.6
f`(x)=5/(2√e^5x)
f`(0)=5/(2√e)


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1.1.f(x)=2·x³/²-3·x²/³+6·x¹/³
 f'(x)=2·(2/3)·x³/²⁻¹-(3·2/3)x²/³⁻¹+6·`1/3·x¹/³⁻¹=(4/3)·x¹/²-2·x⁻¹/³+2·x⁻²/³=
=4√x/3-2/∛x+2/∛x²
f'(1)=4/3-2+2=4/3.
1.2.f(x)=2x/(1+x²).
    f'(x)=2(1+x²)-2x·2x)/(1+x²)²=2+2x²-4x²)/( 1+x²)²=(2-2x²)/(1+x²)²
    f'(-3)=(2-2·(-3)²)/(1(-3)²)²=(2-18)/10²=-16/100=-0.16
1.3.f(x)=(x-1)·√(x²+1)
      f'(x)=√(x²+1)+(x-1)·1/2·√(x²+1)·2x=√(x²+1)+x(x-1)/·(x²+1)=
=√(x²+1)+(1+x²-x)/ √(x²+1) 
f'(0)=(√(0+1)+1+0-0)/√1)=1+1=2
1.4.f(x)=cos³x/3-cosx
  f'(x)=3cos²x·(-sinx)/3+sinx=sinx(1-cos²x)=sinx·sin²x=sin³x
  f'(π/6)=(1/2)³=1/8=0.125
1.5.f(x)=4ln(x-1)/x+1)
f'(x)=4/(x-1)/(x+1)·((x+1)-(x-1))/(x+1)²)=(4(x+1)/(x-1))·2/(x+1)²=8/(x²-1)
f'(-3)=8/((-3)²-1)=8/8=1
1.6.f(x)=√e⁵ˣ
     f'(x)=1/2√e⁵ˣ ·e⁵ˣ·5=5√e⁵ˣ/3
    f'(0)=5·e⁰/2=5/2=2.5
2.a(t)=v'(t)
  a(t)=4t-3
  a(3)=4·3-3=9(м/с²)
  4t-3=0,4t=3,t=3/4=0,75c

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