Question

If \( x^2 + y^2 = 1 \), then

• A. \( yy'' + (y')^2 + 1 = 0 \)
• B. \( yy'' + 2(y')^2 + 1 = 0 \)
• C. \( yy'' - 2(y')^2 + 1 = 0 \)
• D. \( yy'' + (y')^2 - 1 = 0 \)
• E. \( yy'' - (2y')^2 - 1 = 0 \)

Hint:

For implicit equations, always differentiate carefully using product rule when variables are mixed.

Answer 1

The Correct Option is A

Concept: We use implicit differentiation twice:
• First derivative: differentiate equation once
• Second derivative: differentiate again

Step 1: Differentiate once
Given: \[ x^2 + y^2 = 1 \] Differentiate w.r.t \(x\): \[ 2x + 2y y' = 0 \] \[ x + y y' = 0 \quad ...(1) \]

Step 2: Differentiate again
Differentiate (1): \[ 1 + \frac{d}{dx}(y y') = 0 \] Use product rule: \[ \frac{d}{dx}(y y') = y' \cdot y' + y y'' \] \[ = (y')^2 + y y'' \] Thus: \[ 1 + (y')^2 + y y'' = 0 \]

Step 3: Rearrange
\[ yy'' + (y')^2 + 1 = 0 \] Final Answer: \[ \boxed{yy'' + (y')^2 + 1 = 0} \]