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

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

tangent of

y = \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}}

\int \frac{8}{( 4 x ^{2} + 1 ) ^{2}} d x

\int \frac{6 - 72 x}{1 - 36 x ^{2}} d x

\frac{d}{d x} (\frac{x}{x + 2})

\frac{d}{d x} (\frac{x + 4}{x - 5})

23 x

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