We have y3−3y+x = 0 Differentiating both sides, we get 3y2y′−3y′+1 = 0 (1) Substituting x = 2√2 , we get y = - 10√2. ⇒ y′(−10√2) = −
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Differentiating Eq. (1), we get 3y2y"+6y(y′)2−3y" = 0 Substituting x = 2√2 , we get y = −10√2 and y' = −