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

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

tangent of

\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}}

\frac{\partial}{\partial y} (sin (x^{2} + x y))

y^{^{''}} - y = 18 t

\frac{d}{d x} (- x^{2} + 6 x)

f (x) = - 5 x^{2}, at x = 6

\frac{\partial}{\partial x} (x - y)

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