What is the tangent of (e^x)/(x+1),\at x=0 ?

The tangent of (e^x)/(x+1),\at x=0 is y=1

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tangent of (e^x)/(x+1),\at x=0

tangent of

\frac{e ^{x}}{x + 1}, at x = 0

y = 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

\int 8 cos^{4} (θ) sin (θ) d θ

\int \frac{5}{16 + x ^{2}} d x

ln (ln (8 s))

f (x) = 4 x^{4} - x^{6}

y = (6 x^{3} + 3 x + 1)^{10}

What is the tangent of (e^x)/(x+1),\at x=0 ?
The tangent of (e^x)/(x+1),\at x=0 is y=1