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}}

Compute the slope of

y = - \frac{6 x}{x ^{2} + 1}

x = a_{0} : 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}}

w \to - 2 lim (\frac{w ^{2} + 5 w + 6}{w ^{3} + 8})

\frac{( e ^{2 x} )}{x ^{2} + 1}

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

\frac{\partial}{\partial x} (x^{\frac{3}{4}})

e^{x} - x

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}