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

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

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

tangent of

f (x) = \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

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

Compute the slope of

f (x) = \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{x}{( 1 - x )})

\int x^{3} x - 4 d x

\frac{d}{d x} (\frac{2 x ^{2} - 3 x + 1}{2 x + 1})

n = 1 \sum \infty ln (1 + \frac{2}{n})

\frac{d}{d x} (3 x^{\frac{1}{4}})

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