What is the tangent of (x^2)/(x+9) ?

The tangent of (x^2)/(x+9) is y=(a_{0}^2+18a_{0})/((a_{0)+9)^2}x-(9a_{0}^2)/((a_{0)+9)^2}

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tangent of (x^2)/(x+9)

tangent of

\frac{x ^{2}}{x + 9}

y = \frac{a _{0}^{2} + 18 a _{0}}{( a _{0} + 9 ) ^{2}} x - \frac{9 a _{0}^{2}}{( a _{0} + 9 ) ^{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} + 9})

Find the slope of

y = \frac{x ^{2}}{x + 9} : \frac{d y}{d x} = \frac{x ^{2} + 18 x}{( x + 9 ) ^{2}}

Compute the slope of

y = \frac{x ^{2}}{x + 9}

x = a_{0} : m = \frac{a _{0}^{2} + 18 a _{0}}{( a _{0} + 9 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} + 18 a _{0}}{( a _{0} + 9 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} + 9}) : y = \frac{a _{0}^{2} + 18 a _{0}}{( a _{0} + 9 ) ^{2}} x - \frac{9 a _{0}^{2}}{( a _{0} + 9 ) ^{2}}

y = \frac{a _{0}^{2} + 18 a _{0}}{( a _{0} + 9 ) ^{2}} x - \frac{9 a _{0}^{2}}{( a _{0} + 9 ) ^{2}}

\int \frac{x ^{3} - 4 x - 10}{x ^{2} - x - 6} d x

\int \frac{3 - 2 x}{5 x ^{2} + 7} d x

\int_{- 1}^{3} (x^{3} - 2) d x

\int - 2 y^{2} d y

x \to 0 lim (\frac{x}{3})

What is the tangent of (x^2)/(x+9) ?
The tangent of (x^2)/(x+9) is y=(a_{0}^2+18a_{0})/((a_{0)+9)^2}x-(9a_{0}^2)/((a_{0)+9)^2}