MHT CET · Maths · Differentiation
If \(x^{2}+y^{2}=1\), then \(\frac{d^{2} x}{d y^{2}}=\)
- A \(x^{3}\)
- B \(y^{3}\)
- C \(-\frac{1}{x^{3}}\)
- D \(-y^{3}\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{x^{3}}\)
Step-by-step Solution
Detailed explanation
We have \(x^{2}+y^{2}=1\)
\(\begin{array}{l}
2 x+2 y \frac{d y}{d x}=0 \Rightarrow 2 y \frac{d y}{d x}=-2 x \\
\frac{d y}{d x}=-\frac{x}{y} \Rightarrow \frac{d x}{d y}=-\frac{y}{x}
\end{array}\)
Differentiating w.r.t. \(\mathrm{y}\) we get
\(\frac{d^{2} x}{d y^{2}}=\left[\frac{x(1)-y \frac{d x}{d y}}{x^{2}}\right]=-\left[\frac{x-y\left(\frac{-y}{x}\right)}{x^{2}}\right]=-\left[\frac{x^{2}+y^{2}}{x^{3}}\right]\)
\(=-\frac{1}{x^{3}} \quad \cdots\left[\because x^{2}+y^{2}=1\right.\), given \(]\)
\(\begin{array}{l}
2 x+2 y \frac{d y}{d x}=0 \Rightarrow 2 y \frac{d y}{d x}=-2 x \\
\frac{d y}{d x}=-\frac{x}{y} \Rightarrow \frac{d x}{d y}=-\frac{y}{x}
\end{array}\)
Differentiating w.r.t. \(\mathrm{y}\) we get
\(\frac{d^{2} x}{d y^{2}}=\left[\frac{x(1)-y \frac{d x}{d y}}{x^{2}}\right]=-\left[\frac{x-y\left(\frac{-y}{x}\right)}{x^{2}}\right]=-\left[\frac{x^{2}+y^{2}}{x^{3}}\right]\)
\(=-\frac{1}{x^{3}} \quad \cdots\left[\because x^{2}+y^{2}=1\right.\), given \(]\)
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