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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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} (e^{x} (x - 5))

2 e^{z}

\frac{d}{d x} (\frac{3 x - 7}{2 x + 4})

f (x) = \frac{7 + tan ( x )}{5 - tan ( x )}

\frac{\partial}{\partial x} (- 3 y cos (3 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}