AP EAMCET · Maths · Differentiation
If \(y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2}+\tan ^{-1} \frac{x}{1+12 x^2}\), then \(\left(\frac{d y}{d x}\right)_{x=\frac{1}{2}}=\)
- A \(1\)
- B \(-1\)
- C \(0\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2}+\tan ^{-1} \frac{x}{1+12 x^2} \\ & \frac{d y}{d x}=\frac{1-2 x^2}{\left(1+2 x^2\right)^2+x^2}+\frac{1-6 x^2}{\left(1+6 x^2\right)^2+x^2}+\frac{1-12 x^2}{\left(1+12 x^2\right)^2+x^2} \\ & \left(\frac{d…
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