If tany= 2^x/1+2^2x+1 then dy/dx at x=0

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y = {tan}^{- 1} (\frac{2^{x^{'}}}{1 + 2^{2 x + 1}})

\frac{d y}{d x}

x = 0

\frac{d y}{d x} - x t a n (y - x) = 1

I f y = {tan}^{- 1} [\frac{2^{x}}{1 + 2^{x + 1}}], - \infty < x < \infty t h e n \frac{d y}{d x} a t x = 0 i s

s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}},

\frac{d y}{d x}

\frac{d y}{d x} + \frac{tan y}{1 + x} = (1 + x) e^{x} sec y .

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