y=tan-12x1-x2 হলে dydx=কত?

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

$y = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ হলে $\frac{d y}{d x}$ =কত?

$- \frac{2}{1 + x^{2}}$

$\frac{2}{1 + x^{2}}$

$\frac{1}{1 + 4 x^{2}}$

$\frac{4}{1 + x^{2}}$

2021

লগইন করুন