曲线y=1-2/(x+2)在点(-1,-1)处的切线方程:2x-y+1=0解:∵y=1-2/(x+2)∴y'=2/(x+2)²∴k=y∣(x=-1)=2∴y+1=2(x+1)即2x-y+1=0为所求。
如果题目是:y = 1 - 2/(x+2) 在(-1,-1)处的切线方程。y = 1 - 2/(x+2)y ' = 2/(x+2)² 当x = -1时 y ' = 2/(-1+2)² = 2所以切线方程: y = 2(x+1)-1 = 2x + 1