What is the tangent of 2sqrt(x)+1 ?

The tangent of 2sqrt(x)+1 is y= 1/(sqrt(a_{0))}x+sqrt(a_{0)}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 2sqrt(x)+1

tangent of

2 x + 1

y = \frac{1}{a _{0}} x + a_{0} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0} + 1)

Find the slope of

y = 2 x + 1 : \frac{d y}{d x} = \frac{1}{x}

Compute the slope of

y = 2 x + 1

x = a_{0} : m = \frac{1}{a _{0}}

Find the line with slope m=

\frac{1}{a _{0}}

and passing through

(a_{0}, 2 a_{0} + 1) : y = \frac{1}{a _{0}} x + a_{0} + 1

y = \frac{1}{a _{0}} x + a_{0} + 1

\int 5^{- 5 x} d x

(x^{9} - 3 x)^{- 2}

\int (y + 1) d y

\frac{d y}{d x} - \frac{x y}{1 + x} = x

\frac{\partial}{\partial x} (ln (4 x^{2} + 9 y^{2} + 8))

What is the tangent of 2sqrt(x)+1 ?
The tangent of 2sqrt(x)+1 is y= 1/(sqrt(a_{0))}x+sqrt(a_{0)}+1