MHT CET · Maths · Inverse Trigonometric Functions
If \(x^{2} y^{2}=\sin ^{-1} \sqrt{x^{2}+y^{2}}+\cos ^{-1} \sqrt{x^{2}+y^{2}}\), then \(\frac{d y}{d x}=\)
- A \(\frac{-y}{x}\)
- B \(\frac{x}{y}\)
- C \(\frac{y}{x}\)
- D \(\frac{-x}{y}\)
Answer & Solution
Correct Answer
(A) \(\frac{-y}{x}\)
Step-by-step Solution
Detailed explanation
(D)
We know, \(\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}\) Differentiating w.r.t. \(\mathrm{x}\)
\(\begin{array}{l}
x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \\
\frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x}
\end{array}\)
We know, \(\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}\) Differentiating w.r.t. \(\mathrm{x}\)
\(\begin{array}{l}
x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \\
\frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x}
\end{array}\)
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