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}}

- 7 x^{4} + 11 x^{2} - 4

\int \frac{x ^{2}}{( 1 + 9 x ^{2} ) ^{\frac{3}{2}}} d x

x \to 6 lim ((x - 2)^{2})

4 (\frac{d y}{d x}) - (y) - x = 0

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

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}