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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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)