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ДженниЛенни
@ДженниЛенни
August 2022
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найти значение производной в точке х0
f(x)=2sinx/2+cos3x,x0=pu/2 - решение.
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Dимасuk
Verified answer
F(x) = 2sinx/(2 + cos3x)
u = sinx, v = 2 + cos3x
f'(x) = (u'v - v'u)/v²
f'(x) = 2[(sinx)'(2 + cos3x) - (2 + cos3x)'sinx]/(2 + cos3x)² = 2[cosx(2 + cos3x) + 3sin3xsinx]/(2 + cos3x)²
f'(π/2) = 2[cos(π/2)·(2 + cos(3π/2)) + 3sin(3π/2)sin(π/2)]/(2 + cos(3π/2)² =
2(0 - 3)/(2 + 0)² = -6/4 = -1,5
Ответ: f'(π/2) = -1,5
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Answers & Comments
Verified answer
F(x) = 2sinx/(2 + cos3x)u = sinx, v = 2 + cos3x
f'(x) = (u'v - v'u)/v²
f'(x) = 2[(sinx)'(2 + cos3x) - (2 + cos3x)'sinx]/(2 + cos3x)² = 2[cosx(2 + cos3x) + 3sin3xsinx]/(2 + cos3x)²
f'(π/2) = 2[cos(π/2)·(2 + cos(3π/2)) + 3sin(3π/2)sin(π/2)]/(2 + cos(3π/2)² =
2(0 - 3)/(2 + 0)² = -6/4 = -1,5
Ответ: f'(π/2) = -1,5