Question

已知y(x)方程sin(xy)+ln(y-x)=x确定,求y'(0)及当x=0时切线方程

Answer 1

对sin(xy)+ln(y-x)=x①求导得
cos(xy)*(y+xy')+(y'-1)/(y-x)=1,
整理得[xcos(xy)+1/(y-x)]y'=1-ycos(xy)+1/(y-x),
∴y'=[1-ycos(xy)+1/(y-x)]/[xcos(xy)+1/(y-x)]
=[y-x+1-(y^2-xy)cos(xy)]/[1+(xy-x^2)cos(xy)],②
把x=0代入①,得lny=0,y=1.
都代入②,y'(0)=1,
∴曲线①在点(0,1)的切线方程是y=x+1.