设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx
来源:学生作业帮 编辑:灵鹊做题网作业帮 分类:数学作业 时间:2024/04/23 15:01:25
设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx
复合函数求导
设 y=f(t),t(x)=(2x-1)/(x+1)
则
dy/dt = lnt^(1/3)=ln{[(2x-1)/(x+1)]^(1/3)},
dt/dx=[(2x-1)/(x+1)]'=3/(x+1)^2
【具体过程】
dy/dx
=(dy/dt)*(dt/dx)
=f'[(2x-1)/(x+1)]*[(2x-1)/(x+1)]'
=ln{[(2x-1)/(x+1)]^(1/3)}*[3/(x+1)^2]
=(1/3)*ln[(2x-1)/(x+1)]*[3/(x+1)^2]
=[ln(2x-1)-ln(x+1)]/(x+1)^2
设 y=f(t),t(x)=(2x-1)/(x+1)
则
dy/dt = lnt^(1/3)=ln{[(2x-1)/(x+1)]^(1/3)},
dt/dx=[(2x-1)/(x+1)]'=3/(x+1)^2
【具体过程】
dy/dx
=(dy/dt)*(dt/dx)
=f'[(2x-1)/(x+1)]*[(2x-1)/(x+1)]'
=ln{[(2x-1)/(x+1)]^(1/3)}*[3/(x+1)^2]
=(1/3)*ln[(2x-1)/(x+1)]*[3/(x+1)^2]
=[ln(2x-1)-ln(x+1)]/(x+1)^2
设y=f((2x-1)/(x+1)),f'(x)=lnx^(1/3),求dy/dx
设y=f(根号lnx),已知dy/dx=1/(2x^2*根号lnx),求f'(x),即f(x)的导数.
设f(x)=1/x,y=f[(x-1)/(x+1)],求dy/dx
y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy
设f(x)为可导函数,求dy/dx:y=f(arcsin(1/x))
设函数Y=f(x)由x2+3y4+x+2y=1所确定,求dy/dx
设f(x)可导,且f'(0=1,又y=f(x^2+sin^2x)+f(arctanx),求dy/dx /x=0
1、 设F(x)=e-x ,求∫f/(lnx)/x dx
设f(x)为可导函数,求dy/dx (1)y=f(tanx) (2)y=f(x^2)+lnf(x)
设y=(x/1-x)^x,求dy/dx
设f(x)= ∫0-x e^(-y+2y)dy 求∫0-1 [(1-x)^2]f(x)dx
设f(x)=lnx,计算不定积分∫(1/ x^2)*f'(1/x)dx