What is the tangent of (x^2)/(x-4) ?

The tangent of (x^2)/(x-4) is y=(a_{0}^2-8a_{0})/((a_{0)-4)^2}x+(4a_{0}^2)/((a_{0)-4)^2}

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tangent of (x^2)/(x-4)

tangent of

\frac{x ^{2}}{x - 4}

y = \frac{a _{0}^{2} - 8 a _{0}}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} - 4 ) ^{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} - 4})

Find the slope of

y = \frac{x ^{2}}{x - 4} : \frac{d y}{d x} = \frac{x ^{2} - 8 x}{( x - 4 ) ^{2}}

Compute the slope of

y = \frac{x ^{2}}{x - 4}

x = a_{0} : m = \frac{a _{0}^{2} - 8 a _{0}}{( a _{0} - 4 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} - 8 a _{0}}{( a _{0} - 4 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} - 4}) : y = \frac{a _{0}^{2} - 8 a _{0}}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} - 4 ) ^{2}}

y = \frac{a _{0}^{2} - 8 a _{0}}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} - 4 ) ^{2}}

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

3^{x}

\int \frac{x ^{2} + 2}{x ^{2} + 1} d x

\int (x + \frac{1}{x} + π) d x

\int \frac{1}{x ^{12}} d x

What is the tangent of (x^2)/(x-4) ?
The tangent of (x^2)/(x-4) is y=(a_{0}^2-8a_{0})/((a_{0)-4)^2}x+(4a_{0}^2)/((a_{0)-4)^2}