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}}

sin (x), \frac{π}{4}

y^{^{''}} + 6 y^{^{'}} + 8 y = 4 sin (2 t)

\int_{1}^{3} x - 1 d x

\frac{d y}{d x} = \frac{e ^{x - y}}{1 + e ^{x}}

\int (1 + 4 x)^{- \frac{1}{2}} d x

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}