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

The tangent of (2x-1)/(x+1) is y= 3/((a_{0)+1)^2}x+(2a_{0}^2-2a_{0}-1)/((a_{0)+1)^2}

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tangent of (2x-1)/(x+1)

tangent of

\frac{2 x - 1}{x + 1}

y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} - 2 a _{0} - 1}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = \frac{2 x - 1}{x + 1} : \frac{d y}{d x} = \frac{3}{( 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{3}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

\frac{3}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{2 a _{0} - 1}{a _{0} + 1}) : y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} - 2 a _{0} - 1}{( a _{0} + 1 ) ^{2}}

y = \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} - 2 a _{0} - 1}{( a _{0} + 1 ) ^{2}}

\frac{v - 1}{v ^{2} - 2 v + 1} d v = - \frac{1}{y} d y

\frac{\partial}{\partial x} (T x^{5} y^{4} z^{8})

\frac{1}{s ^{2} + 5 s + 2}

(\frac{3 x + 2}{x - 1})^{7}

(2 x^{3} + 7)^{2}

What is the tangent of (2x-1)/(x+1) ?
The tangent of (2x-1)/(x+1) is y= 3/((a_{0)+1)^2}x+(2a_{0}^2-2a_{0}-1)/((a_{0)+1)^2}