Question

设y=y（x） 由方程x^3+y^3=e^xy 确定,求 dy/dx!x=0的值

Answer 1

x^3+y^3=e^xy

对x求导

3x²+3y²*y'=e^(xy)*(xy)'

3x²+3y²*y'=e^(xy)*(y+x*y')

3x²+3y²*y'=e^(xy)*y+e^(xy)*x*y'

y'=[3x²-e^(xy)*y]/[e^(xy)*x-3y²]

即dy/dx=[3x²-e^(xy)*y]/[e^(xy)*x-3y²]

x=0,代入x^3+y^3=e^xy

0+y^3=1

y=1

代入dy/dx=[3x²-e^(xy)*y]/[e^(xy)*x-3y²]

所以dy/dx=(0-1)/(0-3)=1/3