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}}

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{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}}

s l o p e (0, - 5), (- 3, 5)

x \to 0 - lim (e^{\frac{4}{x}})

\frac{\partial}{\partial y} (x + y e^{\frac{y}{x}})

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

\frac{\partial}{\partial x} (9 x sin (5 x^{2} y))

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}