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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 1/(x+2)

tangent of

\frac{1}{x + 2}

y = - \frac{1}{( a _{0} + 2 ) ^{2}} x + \frac{2 a _{0} + 2}{( a _{0} + 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} + 2})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = - \frac{1}{( a _{0} + 2 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} + 2 ) ^{2}}

and passing through

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

y = - \frac{1}{( a _{0} + 2 ) ^{2}} x + \frac{2 a _{0} + 2}{( a _{0} + 2 ) ^{2}}

\int_{0}^{\frac{π}{3}} sec (x) d x

\int \frac{x}{5 x + 1} d x

2 x + 7

(2 x y^{2} - 5) d x + (2 x^{2} y + 4) d y = 0

x \to 1 lim (x^{4} - 1)

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