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

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

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

tangent of

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

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

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

y = \frac{2}{3} x^{3} - \frac{3}{4 x ^{2}}

\frac{\partial}{\partial y} (\frac{cos ( y )}{x z ^{2}})

y^{^{'''}} = 6

\int sin (3 π t) d t

\int \frac{1}{5 - 2 x} d x

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