论文部分内容阅读
求曲线y=3x-χ~3过点A(2,-2)的切线方程.错解:显然点A在曲线y=3χ-χ~3上.设f(χ)=3χ-χ~3,则f’(χ)=3-3χ~2,故f’(2)=-9.
Find the curve y = 3x-χ ~ 3 point A (2, -2) of the tangent equation. Misinterpretation: Obviously point A in the curve y = 3χ-χ ~ 3. Let f (χ) = 3χ-χ ~ 3 , Then f ’(χ) = 3-3χ ~ 2, so f’ (2) = -9.