What is the tangent of f(x)= x/(x^2+9) ?

The tangent of f(x)= x/(x^2+9) is y=(-a_{0}^2+9)/((a_{0)^2+9)^2}x+(2a_{0}^3)/((a_{0)^2+9)^2}

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tangent of f(x)= x/(x^2+9)

tangent of

f (x) = \frac{x}{x ^{2} + 9}

y = \frac{- a _{0}^{2} + 9}{( a _{0}^{2} + 9 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 9 ) ^{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} + 9})

Find the slope of

f (x) = \frac{x}{x ^{2} + 9} : \frac{df ( x )}{d x} = \frac{- x ^{2} + 9}{( x ^{2} + 9 ) ^{2}}

Compute the slope of

f (x) = \frac{x}{x ^{2} + 9}

x = a_{0} : m = \frac{- a _{0}^{2} + 9}{( a _{0}^{2} + 9 ) ^{2}}

Find the line with slope m=

\frac{- a _{0}^{2} + 9}{( a _{0}^{2} + 9 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{a _{0}^{2} + 9}) : y = \frac{- a _{0}^{2} + 9}{( a _{0}^{2} + 9 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 9 ) ^{2}}

y = \frac{- a _{0}^{2} + 9}{( a _{0}^{2} + 9 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 9 ) ^{2}}

x (x + y) y^{^{'}} + y (3 x + y) = 0

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2 t y^{^{'}} + y = e^{t}

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\int_{0}^{27} \frac{4}{x ^{\frac{2}{3}}} d x

What is the tangent of f(x)= x/(x^2+9) ?
The tangent of f(x)= x/(x^2+9) is y=(-a_{0}^2+9)/((a_{0)^2+9)^2}x+(2a_{0}^3)/((a_{0)^2+9)^2}