f(x)=定积分上限为x^2,下限为0的cos2tdt,则f'(根号下pai/x）=

Question

如题所述

Answer 1

f(x)=∫(x^2,0)cos2tdt =∫(x^2,0)dsin2t * (1/2)=(sin2x^2)/2;

f'(根号下pai/x）=f'(sin(pai/x)) /2 =cos(pai/x) *pai / 2

Answer 2

f(x) = ∫(0，x²) cos(2t) dt
f'(x) = 2xcos(2x²) - 0
f'(√(π/x)) = 2√(π/x)cos[2√(π/x)²]
= 2√(π/x)cos(2π/x)

Answer 3

f==sin(2*x^2)/2
f'=2*x*cos(2*x^2);
f'(根号下pai/x）=(2*pi*cos((2*pi^2)/x^2))/x追问

没看懂.....

追答

不是吧！汗~
f(x)=定积分上限为x^2,积分结果：f=sin(2*x^2)/2；对f求导：f'=2*x*cos(2*x^2);
要求f'(根号下pai/x）=？将根号下pai/x代入f‘表达式，结果：f'(根号下pai/x）=(2*pi*cos((2*pi^2)/x^2))/x

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