设y=sinx/x^2,求f'(π/3)
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设y=sinx/x^2,求f''(π/3)设y=sinx/x^2,求f''(π/3)设y=sinx/x^2,求f''(π/3)已知f(x)=(sinx)/x²,求f''(π/3).f''(x)=(x&
设y=sinx/x^2,求f'(π/3)
设y=sinx/x^2,求f'(π/3)
设y=sinx/x^2,求f'(π/3)
已知f(x)=(sinx)/x²,求f '(π/3).
f '(x)=(x²cosx-2xsinx)/x⁴=(xcosx-2sinx)/x³;
故f '(π/3)=[(π/3)(1/2)-2(√3/2)]/(π/3)³=27[(π/6)-√3]/π³=[(9π/2)-27√3]/π³=9/(2π²)-(27√3)/π³
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