y = $\tan^{-1} \frac{4x}{1 - 4x^2}$ হলে $\frac{dy}{dx} = ?$ (ক) $\frac{2}{1 + 4x^2}$ (খ) $\frac{2}{1 - 4x^2}$ (গ) $\frac{4}{1 - 4x^2}$ (ঘ) $\frac{4}{1 + 4x^2}$

ID#7038 HSC Higher Math 1st MCQ (Barisal 2023)

y = $\tan^{-1} \frac{4x}{1 - 4x^2}$ হলে $\frac{dy}{dx} = ?$

ক) (ক) $\frac{2}{1 + 4x^2}$

খ) (খ) $\frac{2}{1 - 4x^2}$

গ) (গ) $\frac{4}{1 - 4x^2}$

ঘ) (ঘ) $\frac{4}{1 + 4x^2}$

ঘ

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