What is the tangent of x/(x^2+4) ?

The tangent of x/(x^2+4) is y=(-a_{0}^2+4)/((a_{0)^2+4)^2}x+(2a_{0}^3)/((a_{0)^2+4)^2}

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tangent of x/(x^2+4)

tangent of

\frac{x}{x ^{2} + 4}

y = \frac{- a _{0}^{2} + 4}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 4 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}}{a _{0}^{2} + 4})

Find the slope of

y = \frac{x}{x ^{2} + 4} : \frac{d y}{d x} = \frac{- x ^{2} + 4}{( x ^{2} + 4 ) ^{2}}

Compute the slope of

y = \frac{x}{x ^{2} + 4}

x = a_{0} : m = \frac{- a _{0}^{2} + 4}{( a _{0}^{2} + 4 ) ^{2}}

Find the line with slope m=

\frac{- a _{0}^{2} + 4}{( a _{0}^{2} + 4 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{a _{0}^{2} + 4}) : y = \frac{- a _{0}^{2} + 4}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 4 ) ^{2}}

y = \frac{- a _{0}^{2} + 4}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 4 ) ^{2}}

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

\frac{1}{s ^{2} + 2 s}

x \to \infty lim (\frac{5}{3})

x \to 0 lim ((3 x + 9)^{x})

\frac{d y}{d x} = - 8 y x

What is the tangent of x/(x^2+4) ?
The tangent of x/(x^2+4) is y=(-a_{0}^2+4)/((a_{0)^2+4)^2}x+(2a_{0}^3)/((a_{0)^2+4)^2}