设y＝tan三次方一÷x求dy dx-搜答案-神马作文网

两边对x求导：y'=(1+y')[sec(x+y)]^2得y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=1/{[cos(x+y)]^2-1}因此dy=dx/{[cos(x+y)]^

ylny=lnx两边对x求导：y'lny+y(1/y)y'=1/xy'lny+y'=1/x(lny+1)y'=1/xy'=1/[x(lny+1)]

左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx

1、两边同时微分,y^3dx+3xy^2dy=dy,sody/dx=(y^3)/(1-3xy^2)2、dy=(e^x)/(1+e^(2x))dx

y=3^x+(cosx)^5dy=[(ln3)3^x+5sinx(cosx)^4]dx

dy=sec²x(dy)^(-x)=(sec²x)^(-x)=sec^(-2x)x

x/y=ln(y/x)x(-1/y^2)y'+1/y=x/y(-y/x^2+y'/x)(1/y+x/y^2)y'=1/y+1/x[(y+x)/y^2]y'=(x+y)/xyy'=y/x

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)

dy/dx=sec²(x+y)*(1+dy/dx)则[1-sec²(x+y)]dy/dx=sec²(x+y)则dy/dx=sec²(x+y)/[1-sec

y=tanx³+2（-x）^2y'=3tanx*sec^2(x)+4xdy=y'dx=[3tanx*sec^2(x)+4x]dx

dy=dlnx+dcos³x=dx/x+3cos²xdcosx=dx/x-3cos²xsinxdx=(1/x-3cos²xsinx)dx

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+1)^2*dy/dx+x^3=0(y+1)^2*dy/dx=-x^3(y+1)^2*dy=-x^3dx1/3*(y+1)^3=-1/4*x^4+C(y+1)^3=-3/4*x^4+C

dy=[-sin(√x)*1/2*x^(-1/2)-e^(-2x)*(-2)]dx=[1/2sin(√x)x^(-1/2)+2e^(-2x)]dx

...y＝ln（x＋e^x^2）dy=（x＋e^x^2）^(-1)*(1+2xe^x^2)x^y就是x的y次方

设y=x*e^y,求dy=?

两边对x求导得y'=e^y+xe^y*y'y'=e^y/(1-xe^y)dy=e^y/(1-xe^y)dx再问：好快....后面的都懂....不过可以说一下为什么两边对x求导后不是e^y+xe^y么.

y=x^3/cosx则y'=dy/dx=(3x^2cosx+x^3sinx)/cos^2x所以dy=(3x^2cosx+x^3sinx)/cos^2xdx

y=x³sinxy'=3x²sinx+x³cosxdy=x²(3sinx+xcosx)dx

第二步确实应该是d（tanx²）=sec²x²d（x²）