What is the tangent of (x^2)/(x+10) ?

The tangent of (x^2)/(x+10) is y=(a_{0}^2+20a_{0})/((a_{0)+10)^2}x-(10a_{0}^2)/((a_{0)+10)^2}

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tangent of (x^2)/(x+10)

tangent of

\frac{x ^{2}}{x + 10}

y = \frac{a _{0}^{2} + 20 a _{0}}{( a _{0} + 10 ) ^{2}} x - \frac{10 a _{0}^{2}}{( a _{0} + 10 ) ^{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} + 10})

Find the slope of

y = \frac{x ^{2}}{x + 10} : \frac{d y}{d x} = \frac{x ^{2} + 20 x}{( x + 10 ) ^{2}}

Compute the slope of

y = \frac{x ^{2}}{x + 10}

x = a_{0} : m = \frac{a _{0}^{2} + 20 a _{0}}{( a _{0} + 10 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} + 20 a _{0}}{( a _{0} + 10 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} + 10}) : y = \frac{a _{0}^{2} + 20 a _{0}}{( a _{0} + 10 ) ^{2}} x - \frac{10 a _{0}^{2}}{( a _{0} + 10 ) ^{2}}

y = \frac{a _{0}^{2} + 20 a _{0}}{( a _{0} + 10 ) ^{2}} x - \frac{10 a _{0}^{2}}{( a _{0} + 10 ) ^{2}}

y = 5^{4 x^{2} + 3 x}

9 x y^{2}

\frac{d}{d t} (ln (t + 1))

f (x) = \frac{2 x}{x + 2}, at x = 2

x y y^{^{'}} = 1 - x^{2}

What is the tangent of (x^2)/(x+10) ?
The tangent of (x^2)/(x+10) is y=(a_{0}^2+20a_{0})/((a_{0)+10)^2}x-(10a_{0}^2)/((a_{0)+10)^2}