What is the tangent of (x^2)/(x+1) ?

The tangent of (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 (x^2)/(x+1)

tangent of

\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

y = \frac{x ^{2}}{x + 1} : \frac{d y}{d x} = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}

Compute the slope of

y = \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}}

x \to 2 lim (9 - x)

\int \frac{2 u}{u + 2} d u

\int 16 x^{3} + \frac{1}{x} d x

\frac{d}{d x} (\frac{3}{4 x ^{3}})

\int_{0}^{3} x^{2} - 9 d x

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