Question

高数第4题，求切线方程

Answer 1

4、
sin(xy)+ln(y-x)=x
cos(xy)(y+xy')+1/(y-x)(y'-1)=1
[xcos(xy)+1/(y-x)]y'=1+1/(y-x)-ycos(xy)
[x(y-x)cos(xy)+1]y'/(y-x)=[(y-x)+1-y(y-x)cos(xy)]/(y-x)
[(xy-x^2)cos(xy)+1]y'=y-x+1-(y^2-xy)cos(xy)
y'=[y-x+1-(y^2-xy)cos(xy)]/[(xy-x^2)cos(xy)+1]
x=0，y=1时
y'=[1-0+1-(1^2-0×1)cos(0×1)]/[0×1-0^2)cos(0×1)+1]
=1
(0,1)处切线方程：y-1=1(x-0)
y=x+1
注：^2——表示平方。