What is the tangent of (-6x)/((x^2+1)) ?

The tangent of (-6x)/((x^2+1)) is y=-(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(12a_{0}^3)/((a_{0)^2+1)^2}

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tangent of (-6x)/((x^2+1))

tangent of

\frac{- 6 x}{( x ^{2} + 1 )}

y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 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{6 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{- 6 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{6 ( - 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{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

- \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, - \frac{6 a _{0}}{a _{0}^{2} + 1}) : y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

What is the tangent of (-6x)/((x^2+1)) ?
The tangent of (-6x)/((x^2+1)) is y=-(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(12a_{0}^3)/((a_{0)^2+1)^2}