If x 2 + y 2 = 1, then

Differentiate x 2 + y 2 = 1 with respect to x : $$2x + 2y\frac{dy}{dx} = 0$$ $$x + y y' = 0$$ Differentiate again: $$1 + (y')^2 + yy'' = 0$$ $$yy'' + (y')^2 = -1$$ or equivalently: $$yy'' + (y')^2 + 1 = 0$$

Limits, Continuity & Differentiability

Second Derivatives

Grade 12

Question:

If x2 + y2 = 1, then

(a) yy'' = 2(y')² + 1 = 0

(b) yy'' + (y')² + 1 = 0

(c) yy'' + (y')² = 1 = 0

(d) yy'' + 2(y')² = 1 = 0

Step-by-Step Solution

Key Concept: Differentiate the constraint x² + y² = 1 twice with respect to x to obtain a relation involving y, y', and y''.

Differentiate x2 + y2 = 1 with respect to x:$$2x + 2y\frac{dy}{dx} = 0$$$$x + y y' = 0$$Differentiate again:$$1 + (y')^2 + yy'' = 0$$$$yy'' + (y')^2 = -1$$or equivalently: $$yy'' + (y')^2 + 1 = 0$$

Correct Answer: C

Question:

Step-by-Step Solution

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