求下列函数的微分dy y=e^tanx-搜答案-搜狗做题网

我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y

y'=e^x(tanx+lnx)+e^x((secx)^2+1/x)=e^x(tanx+lnx+(secx)^2+1/x)dy=[e^x(tanx+lnx+(secx)^2+1/x)]dx

y=e^xcosx+根号2y'=e^xcosx+e^x(-sinx)y'=e^x(cosx-sinx)y=e^(-2x)dy/dx=-2e^(-2x)dy=-2e^(-2x)dxy=1-xe^xdy/

(1)y=3x^2-ln1/x=3x^2+lnxdy=6xdx+(1/x)dx=(6x+1/x)dx(2)y=e^(-x)cosxdy=-e^(-x)cosxdx-e^(-x)sinxdx=-e^(-

先求导：y‘=e^x-3cos3xx=1时,dy=y‘（1）dx=(e-3cos3)dx

dy=（2xsinx+x的平方cosx+2*e的2x次方）dx

dy=e^(x^x)(e^(xlnx))'dx=e^(x^x)*(x^x)*(1+lnx)

两边同时微分得2xdy+2ydx+dy=0(2x+1)dy=-2ydxdy/dx=-2y/(2x+1)(2x+1)y=1y=1/(2x+1)dy/dx=-2/(2x+1)^2

dy=arcsinxdx+xdx/根号(1-x^2)+xdx/(根号1-x^2+e^2)

（1）y=ln(3x^2+e^2),求dy/dx;dy/dx=6x/[3x²+e²]（2）y=∫0^xcost^2dt,求dy/dx;=∫(1/2)(1+cos2t)=(1/2)t

额,自己看看数分书领悟吧.自己弄懂了才能以不变应万变.

再答：��Ƭ�е�ģ��ϣ��>_

解y'=(e^sinx²)'=e^sinx²(sinx²)'=cosx²e^sinx²×(x²)'=2xcosx²e^(sinx&

dy＝y＇dx

y'=[sec(e^x)]^2*(e^x)'=e^x*[sec(e^x)]^2所以dy=e^x*[sec(e^x)]^2dx

直接求导：e^y*y'+y+x*y'=2y*y'解得y'=dy/dx=(-y)/(e^y+x-2y).

dy=[-cscx^2-cscx*cotx]dx

y'=[(lnx)'sinx-lnx*(sinx)']/(sinx)^2=(sinx*1/x-lnx*cosx)/(sinx)^2所以dy=(sinx*1/x-lnx*cosx)/(sinx)^2dx