MHT CET · Maths · Differentiation
If \(\mathrm{y}=\tan ^{-1}\left(\frac{12 x-64 x^3}{1-48 x^2}\right)\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}=\)
- A \(\frac{3}{1+16 x^2}\)
- B \(\frac{4}{1+16 x^2}\)
- C \(\frac{12}{1+16 x^2}\)
- D \(\frac{1}{1+16 x^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{12}{1+16 x^2}\)
Step-by-step Solution
Detailed explanation
Let \(4x = \tan\theta\). \(y = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}\right) = \tan^{-1}(\tan(3\theta)) = 3\theta\)
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