What is the tangent of f(x)=(x-1)/(x+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(x-1)/(x+1)

tangent of

f (x) = \frac{x - 1}{x + 1}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = \frac{x - 1}{x + 1}

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

Find the line with slope m=

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

and passing through

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

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

\frac{d}{d x} (7 - 5 x)

x \to 0 lim (x^{2} + 2)

\int \frac{10}{e ^{\frac{x}{4}}} d x

\int (sin (2 x + y) - sin (x + y)) d x

\frac{d}{d s} (s^{2} y (s))

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