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

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

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

tangent of

\frac{6 x}{2 + x ^{2}}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6 a _{0}}{2 + a _{0}^{2}})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = \frac{6 ( 2 - a _{0}^{2} )}{( 2 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

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

f (x) = (2 x + 3)^{11} (4 - 5 x)^{9}

x \to 1 + lim (1)

\frac{9}{10} x^{2} d y - 18 d x = 0

\frac{\partial}{\partial x} (x^{3} + x^{2} y - y^{2} - 4 y)

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