Question

If \( x^2 + y^2 = 1 \), then find \( \dfrac{d^2x}{dy^2} \).

• A. \( x^3 \)
• B. \( y^3 \)
• C. \( -\dfrac{1}{x^3} \)
• D. \( -y^3 \)

Hint:

For second derivatives in implicit relations, always substitute the original equation to simplify the final expression.

Answer 1

The Correct Option is C

Step 1: Differentiate implicitly with respect to \( y \).
\[ x^2 + y^2 = 1 \] \[ 2x\frac{dx}{dy} + 2y = 0 \] \[ \frac{dx}{dy} = -\frac{y}{x} \]

Step 2: Differentiate again with respect to \( y \).
\[ \frac{d^2x}{dy^2} = -\frac{d}{dy}\!\left(\frac{y}{x}\right) \] Using quotient rule, \[ \frac{d^2x}{dy^2} = -\frac{x(1)-y\frac{dx}{dy}}{x^2} \]

Step 3: Substitute \( \frac{dx}{dy} = -\frac{y}{x} \).
\[ \frac{d^2x}{dy^2} = -\frac{x+\frac{y^2}{x}}{x^2} = -\frac{x^2+y^2}{x^3} \]

Step 4: Use the given relation.
\[ x^2+y^2=1 \Rightarrow \frac{d^2x}{dy^2}=-\frac{1}{x^3} \]

Step 5: Conclusion.
\[ \boxed{-\dfrac{1}{x^3}} \]