What is the tangent of (x^2)/(x-1) ?

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

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tangent of (x^2)/(x-1)

tangent of

\frac{x ^{2}}{x - 1}

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

Find the slope of

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

Compute the slope of

y = \frac{x ^{2}}{x - 1}

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

Find the line with slope m=

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

and passing through

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

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

f (x) = \frac{1}{ln ( x )}

\frac{d}{d x} (\frac{3 - x e ^{x}}{x + e ^{x}})

\frac{d}{d x} (cos (2 x) sin (3 x))

\int (7 x^{6} - \frac{6}{x ^{6}}) d x

\int x^{2} - 3 x + 8 d x

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