What is the tangent of (x^2)/(x+5) ?

The tangent of (x^2)/(x+5) is y=(a_{0}^2+10a_{0})/((a_{0)+5)^2}x-(5a_{0}^2)/((a_{0)+5)^2}

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tangent of (x^2)/(x+5)

tangent of

\frac{x ^{2}}{x + 5}

y = \frac{a _{0}^{2} + 10 a _{0}}{( a _{0} + 5 ) ^{2}} x - \frac{5 a _{0}^{2}}{( a _{0} + 5 ) ^{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} + 5})

Find the slope of

y = \frac{x ^{2}}{x + 5} : \frac{d y}{d x} = \frac{x ^{2} + 10 x}{( x + 5 ) ^{2}}

Compute the slope of

y = \frac{x ^{2}}{x + 5}

x = a_{0} : m = \frac{a _{0}^{2} + 10 a _{0}}{( a _{0} + 5 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} + 10 a _{0}}{( a _{0} + 5 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} + 5}) : y = \frac{a _{0}^{2} + 10 a _{0}}{( a _{0} + 5 ) ^{2}} x - \frac{5 a _{0}^{2}}{( a _{0} + 5 ) ^{2}}

y = \frac{a _{0}^{2} + 10 a _{0}}{( a _{0} + 5 ) ^{2}} x - \frac{5 a _{0}^{2}}{( a _{0} + 5 ) ^{2}}

\int_{- 5}^{6} 0

\frac{d}{d x} (- 6 sin (x) - 6 x cos (x) + x^{2} sin (x))

y = x^{8} f (x)

\int (1 + x)^{5} d x

x \to \infty lim (\frac{5 x}{5 x + 3 x ^{\frac{1}{x}}})

What is the tangent of (x^2)/(x+5) ?
The tangent of (x^2)/(x+5) is y=(a_{0}^2+10a_{0})/((a_{0)+5)^2}x-(5a_{0}^2)/((a_{0)+5)^2}