已知曲线y=y(x)满足siny+x·ey=0,试求曲线在点(0,0)处的切线方程。
1个回答
关注
展开全部
y=y(x)满足siny+xe∧y=0,将方程转为x=-siny/e∧y (A)对(A)求导得x‘ =[-cosy+siny]/e∧y (求导过程不写了)当y=0时代入求导式,x‘ =x/y(定义得来) = -1现在是把x,与y对调,也就是把y当自变量但要求y‘ =y/x = -1
咨询记录 · 回答于2022-12-24
已知曲线y=y(x)满足siny+x·ey=0,试求曲线在点(0,0)处的切线方程。
已知曲线y=y(x)满足siny+x·ey=0,试求曲线在点(0,0)处的切线方程是y=-x+1
y=y(x)满足siny+xe∧y=0,将方程转为x=-siny/e∧y (A)对(A)求导得x‘ =[-cosy+siny]/e∧y (求导过程不写了)当y=0时代入求导式,x‘ =x/y(定义得来) = -1现在是把x,与y对调,也就是把y当自变量但要求y‘ =y/x = -1