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}}

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

\int \frac{1}{y ^{2} + y + 1} d y

\frac{d}{d x} (e^{l n (c o s (x^{2}))} 2 x + 1)

\frac{d y}{d x} + 2 x y = 5 x

(- 4, - 6), (- 2, 0)

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}