What is the tangent of y=(e^{-x})/(x+1),\at x=1 ?

The tangent of y=(e^{-x})/(x+1),\at x=1 is y=-3/(4e)x+5/(4e)

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tangent of y=(e^{-x})/(x+1),\at x=1

tangent of

y = \frac{e ^{- x}}{x + 1}, at x = 1

y = - \frac{3}{4 e} x + \frac{5}{4 e}

Solution steps

Find the tangent point:

(1, \frac{1}{2 e})

Find the slope of

y = \frac{e ^{- x}}{x + 1} : \frac{d y}{d x} = \frac{- e ^{- x} x - 2 e ^{- x}}{( x + 1 ) ^{2}}

Compute the slope of

y = \frac{e ^{- x}}{x + 1}

x = 1 : m = - \frac{3}{4 e}

Find the line with slope m=

- \frac{3}{4 e}

and passing through

(1, \frac{1}{2 e}) : y = - \frac{3}{4 e} x + \frac{5}{4 e}

y = - \frac{3}{4 e} x + \frac{5}{4 e}

\frac{d}{d x} (\frac{2 x + 1}{x ^{2} - 1})

x^{2} - 1, - x + 2, x = 0 x = 1

x^{4} + 9 e^{x}

\frac{d y}{d x}, y = e^{x}

\frac{d y}{d t} - \frac{2 t}{1 + t ^{2}} y = 3

What is the tangent of y=(e^{-x})/(x+1),\at x=1 ?
The tangent of y=(e^{-x})/(x+1),\at x=1 is y=-3/(4e)x+5/(4e)