What is the tangent of x/(x-2) ?

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

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tangent of x/(x-2)

tangent of

\frac{x}{x - 2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}}{a _{0} - 2})

Find the slope of

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

Compute the slope of

y = \frac{x}{x - 2}

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

Find the line with slope m=

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

and passing through

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

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

\int cos (x) \cdot sin (x) d x

\int x cos^{4} (x^{2}) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 2 t}

x^{'} (t) = \frac{1}{3 x ^{2} + 1}, x (0) = 1

\frac{\partial}{\partial x} (e^{- 6 t} cos (π x))

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