设函数y=y(x)有方程∫e^t^2dt(积分从0到y)+∫cos根号下tdt(积分从x^2到1)=0(x>0),求dy⼀dx。

现在急求!要详细过程!
2025-01-08 13:12:25
推荐回答(1个)
回答1:

let
dF(x) = e^(x^2)dx
dG(x) = cos√xdx
∫(0->y)e^t^2dt+∫(x^2->1) cos√tdt =0
F(y) -F(0) + G(1) - G(x^2) =0
d/dx {F(y) -F(0) + G(1) - G(x^2) } =0
F'(y) dy/dx - 2xG'(x^2) =0
e^(y^2) dy/dx - 2xcosx =0
dy/dx = 2xcosx /e^(y^2)