MHT CET · Maths · Differentiation
If \(x y=\tan ^{-1}(x y)+\cot ^{-1}(x y)\), then \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(4,2)}=(\) where \(x, y \in I R)\)
- A \(\frac{-1}{2}\)
- B -2
- C 2
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{-1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& x y=\tan ^{-1}(x y)+\cot ^{-1}(x y) \\
& \Rightarrow x y=\frac{\pi}{2}
\end{aligned}\)
diff. we get \(1 \cdot y+x \cdot \frac{d y}{d x}=0\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{x} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at, }(4,2)=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}\)
& x y=\tan ^{-1}(x y)+\cot ^{-1}(x y) \\
& \Rightarrow x y=\frac{\pi}{2}
\end{aligned}\)
diff. we get \(1 \cdot y+x \cdot \frac{d y}{d x}=0\)
\(\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{x} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at, }(4,2)=\frac{-2}{4}=\frac{-1}{2}
\end{aligned}\)
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