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

The tangent of (2x)/((x^2+1)) is y=(2(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(4a_{0}^3)/((a_{0)^2+1)^2}

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

tangent of

\frac{2 x}{( x ^{2} + 1 )}

y = \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{4 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2 a _{0}}{a _{0}^{2} + 1})

Find the slope of

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

Compute the slope of

y = \frac{2 x}{x ^{2} + 1}

x = a_{0} : m = \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{2 a _{0}}{a _{0}^{2} + 1}) : y = \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{4 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{4 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

x \to - 1 lim (2 x^{2} + 3 x + 1)

\int 21 y d y

y^{^{''}} - \frac{2}{r} y^{^{'}} + \frac{2}{r ^{2}} y = 0, y (1) = 1, y (2) = 0

\int 12 x^{2} - 48 x + 36 d x

x \to 3 lim (\frac{3 x ^{2} - 4 x - 15}{x ^{2} - 5 x + 6})

What is the tangent of (2x)/((x^2+1)) ?
The tangent of (2x)/((x^2+1)) is y=(2(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(4a_{0}^3)/((a_{0)^2+1)^2}