MHT CET · Maths · Differentiation
If \(\frac{x^2}{a^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), then \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}\) is
- A \(\frac{-\mathrm{b}^4}{a}\)
- B \(\frac{b^4}{a^2}\)
- C \(\frac{-\mathrm{b}^4}{\mathrm{y}^3}\)
- D \(\frac{-b^4}{a^2 \cdot y^3}\)
Answer & Solution
Correct Answer
(D) \(\frac{-b^4}{a^2 \cdot y^3}\)
Step-by-step Solution
Detailed explanation
\(\frac{2x}{a^2} + \frac{2y}{b^2}\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{b^2 x}{a^2 y}\)
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