y=x*x (2x-1)求dy-搜答案-智慧作业帮

dy/dx,y=(1+x+x^2)e^x

dy/dx=(1+x+x²)'*e^x+(1+x+x²)*(e^x)'=(1+2x)e^x+(1+x+x²)e^x=(2+3x+x²)e^x

y=(1+x^4)^cos x,求dy

(1+X4)COSx先求导数再乘以dx就行了

sin(x^2+y^2)=x,求dy

sin(x^2+y^2)=x两边同时求导,得(x^2+y^2)'cos(x^2+y^2)=dx(2xdx+2ydy)cos(x^2+y^2)=dx2xdx+2ydy=dx/cos(x^2+y^2)2y

y = x^2 sin 4x,求dy,

y'=2xsin4x-x²cos4x·4所以dy=(2xsin4x-4x²cos4x)dxy=ln√4＋t²=1/2ln(4+t²)y'=1/2·1/(4+t&

y=2x^2+x求dy/dx

dy=d2x²++dxdy=2xdx+dx所以dy/dx=2x+1

y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy两边对x求导：dy/dx=f'[(x-1)/(x+1)]*2/（x+1)^2=arctan[(x-1)/(x+1)]

lny=xln(2+x)dlny=dxln(2+x)dy/y=ln(2-x)dx+x*1/(2+x)dxdy/(2+x)^x=[ln(2-x)+x/(2+x)]dxdy=(2+x)^x[ln(2-x)

y=ln(x+x^2) 求dy

y'=1/(x+x^2)*(2x+1)=(2x+1)/(x+x^2)dy=(2x+1)/(x+x^2)dx

y=ln(x/(1+x))-cot2xdy=[(1+x)/x]d(x/(1+x))+(csc2x)^2.d(2x)={(1+x)/[x(1+x)^2]+2(csc2x)^2}dx

y=x^2+3x+1 求dy

y=x^2+3x+1dy=2x+3再问：��˼·再答：��

y=x^2乘2^x 求dy

先两边取ln,得到lny=xln(x/1-x),然后两边求导,（dy/dx)*(1/y)=ln(x/1-x)+1/1-x.最后只要两边同乘y,把y用题目中的式子代进去就行了.

y=sin^2 {x/(x+1)} 求dy

dy=2sin[x(x+1)]cos[x(x+1)](2x+1)

y=1/x+ln根x.求dy

dy/dx=(x-y+5)/(x+y-2)=[(x+3/2)-(y-7/2)]/[(x+3/2)+(y-7/2)]令v=y-7/2,u=x+3/2,原方程化为dv/du=(u-v)/(u+v)变为齐次

y=arctan(1-x/1+x),求dy

此题复合求导dy=d[arctan(1-x/1+x)]=[1/(1+(1-x/1+x)^2)]·(1-x/1+x)';注：(arctanx)'=1/(1+x^2)=-(1/(x^2+1))

设y=sin[x/(1=x)]求dy

1=x?

这种函数求导,就是一步步求下去就可以的应该就是这样做下去就行了

y=x^2x,求dy

y=x^(2x)lny=2xlnx(1/y)dy=(2+2lnx)dxdy=x^(2x).(2+2lnx)dx

y=e^2x+x^2-yarcsin(1／x)y'=2e^2x+2x-y'arcsin(1／x)+y1/[x^2*√(1-1/x^2)]y'={2e^2x+2x+y1/[x^2*√(1-1/x^2)]