设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx

设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx 数学题求dy/dx设 f'(x)=sin√x 定义（x>0）,又y=f[e^(2x)]求dy/dx 一道微积分题目设f(x)=∫(0到x)e^(-y^2+2y)dy求∫(0到1)((x-1)^2)f(x)dx要过程不要只 设f(x)可导,且f'(0=1,又y=f(x^2+sin^2x)+f(arctanx),求dy/dx /x=0 y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy 1、设函数y=y(x)由方程e^x-e^y=sin(xy)所确定,求(dy/dx)|x=0；2、设函数f(x)=x^2+ 设f(x)=1/x,y=f[(x-1)/(x+1)],求dy/dx 交换积分次序∫(1,0)dx∫(x,0)f(x,y)dy+∫(2,1)dx∫(2-x,0)f(x,y)dy 设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx 设曲线f(x)在[0,1]上可导,且y=f(sin^2x)+f(cos^2x),求dy/dx 设f(u)为可导函数,求dy/dx:(1) y=f(x^3) ; (2) y=f(e^x+x^e); (3) y=f(e 设f(x)连续,Y=∫0~X tf（x^2-t^2）dt 则dy/dx=?