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

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

tangent of

(\frac{2 x - 1}{x + 1}), 2

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}}

Compute the slope of

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

x = a_{0} : 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}}

\int 8 arctan (x) d x

z \to 2 lim ((- \frac{2}{z + 2})^{3})

\frac{d y}{d x} = e^{2 x} + 3 y

\int x 16 + x^{2} d x

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

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