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}}

Compute the slope of

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

x = a_{0} : 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}}

\int_{- \infty}^{3} \frac{1}{4 + x ^{2}} d x

f (x) = e^{9 x} + 9 e^{9 x} x

\frac{d}{d x} (x^{3} + x e^{y} - 2 y)

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

\int 6 x (- x^{2} + 5)^{7} 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}