MHT CET · Maths · Differentiation
If \(x^2 y^2=\sin ^{-1} x+\cos ^{-1} x\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=1\) and \(y=2\) is
- A \(\frac{1}{2}\)
- B 2
- C \(-\frac{1}{2}\)
- D -2
Answer & Solution
Correct Answer
(D) -2
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& x^2 y^2 \\
& = \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\end{aligned}\)
Differentiating w.r.t. x, we get
\(\begin{aligned}
& 2 x y^2+2 y x^2 \frac{d y}{d x}=0 \\
\therefore & \left.\frac{\mathrm{~d} y}{d x}\right|_{(1,2)}=-2
\end{aligned}\)
& x^2 y^2 \\
& = \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\end{aligned}\)
Differentiating w.r.t. x, we get
\(\begin{aligned}
& 2 x y^2+2 y x^2 \frac{d y}{d x}=0 \\
\therefore & \left.\frac{\mathrm{~d} y}{d x}\right|_{(1,2)}=-2
\end{aligned}\)
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