What is the tangent of x/(x^2+49) ?

The tangent of x/(x^2+49) is y=(-a_{0}^2+49)/((a_{0)^2+49)^2}x+(2a_{0}^3)/((a_{0)^2+49)^2}

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tangent of x/(x^2+49)

tangent of

\frac{x}{x ^{2} + 49}

y = \frac{- a _{0}^{2} + 49}{( a _{0}^{2} + 49 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 49 ) ^{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} + 49})

Find the slope of

y = \frac{x}{x ^{2} + 49} : \frac{d y}{d x} = \frac{- x ^{2} + 49}{( x ^{2} + 49 ) ^{2}}

Compute the slope of

y = \frac{x}{x ^{2} + 49}

x = a_{0} : m = \frac{- a _{0}^{2} + 49}{( a _{0}^{2} + 49 ) ^{2}}

Find the line with slope m=

\frac{- a _{0}^{2} + 49}{( a _{0}^{2} + 49 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{a _{0}^{2} + 49}) : y = \frac{- a _{0}^{2} + 49}{( a _{0}^{2} + 49 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 49 ) ^{2}}

y = \frac{- a _{0}^{2} + 49}{( a _{0}^{2} + 49 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 49 ) ^{2}}

t \to 2 lim (\frac{t ^{2} - 4}{t - 2})

x^{^{'}} + 4 x = t e^{- 4 t}

\frac{d}{d x} (e^{y} sin (x y))

x \to - \infty lim (2 x - x^{2})

\int 2 x x^{2} - 1 d x

What is the tangent of x/(x^2+49) ?
The tangent of x/(x^2+49) is y=(-a_{0}^2+49)/((a_{0)^2+49)^2}x+(2a_{0}^3)/((a_{0)^2+49)^2}