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

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

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

tangent of

y = \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}}

\frac{\partial}{\partial x} (\frac{1}{x ^{2}} + y)

\frac{\partial}{\partial y} (y^{2})

y^{^{''}} + y^{^{'}} - 2 y = 2 t, y (0) = 0, y^{^{'}} (0) = 4

\frac{\partial}{\partial x} (sin (x) cosh (y))

\frac{d y}{d x} = 1 + x + y + x \cdot y

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