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

The tangent of y=((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 y=((x-1))/(x+1)

tangent of

y = \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}}

- \frac{5}{7 x + 5}

x \to 0 + lim ((1 + \frac{1}{x})^{x})

\int sin (π t) d t

x \to \infty lim (\frac{x}{e})

\int cos^{3} (\frac{x}{20}) d x

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