What is the tangent of x/(x^2+16) ?

The tangent of x/(x^2+16) is y=(-a_{0}^2+16)/((a_{0)^2+16)^2}x+(2a_{0}^3)/((a_{0)^2+16)^2}

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tangent of x/(x^2+16)

tangent of

\frac{x}{x ^{2} + 16}

y = \frac{- a _{0}^{2} + 16}{( a _{0}^{2} + 16 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 16 ) ^{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} + 16})

Find the slope of

y = \frac{x}{x ^{2} + 16} : \frac{d y}{d x} = \frac{- x ^{2} + 16}{( x ^{2} + 16 ) ^{2}}

Compute the slope of

y = \frac{x}{x ^{2} + 16}

x = a_{0} : m = \frac{- a _{0}^{2} + 16}{( a _{0}^{2} + 16 ) ^{2}}

Find the line with slope m=

\frac{- a _{0}^{2} + 16}{( a _{0}^{2} + 16 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{a _{0}^{2} + 16}) : y = \frac{- a _{0}^{2} + 16}{( a _{0}^{2} + 16 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 16 ) ^{2}}

y = \frac{- a _{0}^{2} + 16}{( a _{0}^{2} + 16 ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 16 ) ^{2}}

y = sin (x^{2})

\int \frac{- 6}{x ^{2} + 16} d x

x \to 1 + lim (\frac{2}{1 - x})

\frac{d}{d x} (a - b)

g (t) = t^{2} - \frac{2}{t ^{3}}

What is the tangent of x/(x^2+16) ?
The tangent of x/(x^2+16) is y=(-a_{0}^2+16)/((a_{0)^2+16)^2}x+(2a_{0}^3)/((a_{0)^2+16)^2}