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

The tangent of f(x)=(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 f(x)=(x^2)/(x-1)

tangent of

f (x) = \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

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

Compute the slope of

f (x) = \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}}

\int 4 x 1 - x^{2} d x

\frac{d}{d x} (\frac{24}{( 1 - x ) ^{4}})

\frac{d}{d x} (\frac{- 1}{x ^{3}})

\frac{4}{5 x + 1}, at x = - 1

x \to 3 lim (3 - x)

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