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

The tangent of e^x+1,\at x=0 is y=x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

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

Compute the slope of

y = e^{x} + 1

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 2) : y = x + 2

y = x + 2

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

\int 2 x^{3} ln (x^{2}) d x

y = 3 + 2 x, y = \frac{9 + x}{3}

\frac{d}{d x} (a^{f (x) (x^{2})})

\frac{\partial}{\partial x} (e^{y} (y^{2} - x^{2}))