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}}

\int_{0}^{l n (2)} s e^{s} d s

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

x \to - 2 lim (- x^{2} + 4 x - 6)

f (x) = x^{2} + 1, (- 2, 5)

\int \frac{x}{( 1 + x ^{3} )} d x

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}