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

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

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tangent of 1/(x-1)

tangent of

\frac{1}{x - 1}

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

Find the slope of

y = \frac{1}{x - 1} : \frac{d y}{d x} = - \frac{1}{( x - 1 ) ^{2}}

Compute the slope of

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

x = a_{0} : m = - \frac{1}{( a _{0} - 1 ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

\int \frac{8 z ^{2} + 2}{2 z ^{3} + z}

\frac{d}{d x} (\frac{e ^{x} cos ( x )}{e ^{x} - 1})

\frac{d}{d t} (4 cos (2 t))

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

\frac{d}{d x} (x (x + 4)^{\frac{1}{3}})

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