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)

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}}

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}

im pl i c i t \frac{d y}{d x}, y = e^{x}

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

d er i v a t i v e tan^{2} (y)

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{2} - 5 x y)

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

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)