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{\partial}{\partial y} (ln (\frac{x y}{y ^{2} + z ^{2}}))

\frac{\partial}{\partial y} (y e^{y x})

3 cos (x) - 2 sin (x)

\int 3 x^{2} + 2 d x

x \to 0 + lim (x - 1)

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}