dy=e∧sinxdx,则y∧n-搜答案-百度作业网

y=e^arcsinx 求dy

y=e^arcsinx求dy=e^（arcsinx）×1/√1-x²dx;如果本题有什么不明白可以追问,

全微分方程通解为(e^x-1)(e^y-1)+c

e^y=sin(x+y)两边求导得e^y*y'=cos(x+y)(x+y)'=cos(x+y)(1+y')=cos(x+y)+y'cos(x+y)[e^y-cos(x+y)]y'=cos(x+y)y'

2e^tan2x*(1/(1+4x^2))-2sin2x*e^cos2x2x*e^x^2对2x+1求积分,得曲线方程为x^2+x+C,又曲线经过（0,1）代入曲线方程得C=1,所以曲线方程为：x^2+

∫(e^y)siny dy=?

∫e^ysinydy=-∫e^yd(cosy)=-[e^y*cosy-∫cosyd(e^y)]=∫cosy*e^ydy-e^ycosy=∫e^yd(siny)-e^ycosy=e^ysiny-∫sin

dy=2[e^x+e^(-x)]*[e^x-e^(-x)]dx再问：��ϸ��再答：��ϸ��ǲ��谡̫��û�취再问：��y��

dy/dx=(dy/dt)/(dx/dt)=(2e^t)′/(3e^-t)′=(2e^t)/(-3e^-t)=-2/3e^2t

y=e^x-ln3ln3是常数的,导数为0dy/dx=e^x

y=e^x^2 则dy=( )

dy=de^(x²)=e^(x²)dx²=2xe^(x²)dx选C

d(e^x+e^y)=dyde^x+de^y=dye^xdx+e^ydy=dy(1-e^y)dy=e^xdxdy/dx=e^x/(1-e^y)

y=sin(e∧x+1),求dy

y=sin(e∧x+1)dy=cos(e^x+1)d(e^x+1)=e^xcos(e^x+1)dx

e∧x+y=xy,求dy/dx

设y=e^(-1/x),则dy=?

dy=（e^(-1/x)）*（-1/x）dx=（e^(-1/x)）*（1/x*x）dx

dy/dx=e∧(x+y）e^(-y)dy=e^xdx两边积分得：-e^(-y)=e^x+ce^y=-1/(e^x+c)y=ln(-1/(e^x+c)y=-ln(c-e^x)

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设y=e^sinx,则dy=

链式法则dy=(e^sinx)*cosxdx

y=x*e^y,则dy=

y=x*e^y,则：y'=e^y+x*e^y*y',所以：y'=e^y/(1-xe^y)=e^y/(1-y)所以：dy={e^y/(1-y)}dx