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

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

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

tangent of

\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

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

Compute the slope of

y = \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}}

\int_{0}^{1} 6 x (1 - x) d x

f (x) = - cos (2 x)

\frac{\partial}{\partial x} (\frac{z - x}{z + y})

\frac{d}{d x} (\frac{16 x}{( 4 - x ^{2} ) ^{2}})

\int \frac{1}{2 x ^{2} + 3 x + 1} d x

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