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vadya11
@vadya11
August 2022
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Вычислите значение производной данной функции в точке X0
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Саша66324
1) f(x) = x²+3x-2, x₀=-1
f'(x)=2x+3
f'(-1)=2*(-1)+3=-2+3=1
ответ: 1
2) f(x)=sinx-cosx, x₀=pi/2
f'(x)=cosx-(-sinx)=cosx+sinx
f'(pi/2)=cos pi/2+sin pi/2=1+0=1
ответ: 1
3)f(x)=x-√x, x₀=4
f'(x)=1+1/(2√x)
f'(4)=1+1/(2√4)=1+1/(2*2)=1+1/4=5/4=1.25
ответ: 1.25
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Answers & Comments
f'(x)=2x+3
f'(-1)=2*(-1)+3=-2+3=1
ответ: 1
2) f(x)=sinx-cosx, x₀=pi/2
f'(x)=cosx-(-sinx)=cosx+sinx
f'(pi/2)=cos pi/2+sin pi/2=1+0=1
ответ: 1
3)f(x)=x-√x, x₀=4
f'(x)=1+1/(2√x)
f'(4)=1+1/(2√4)=1+1/(2*2)=1+1/4=5/4=1.25
ответ: 1.25