What is the tangent of 8/((x^2+4)) ?

The tangent of 8/((x^2+4)) is y=-(16a_{0})/((a_{0)^2+4)^2}x+(24a_{0}^2+32)/((a_{0)^2+4)^2}

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tangent of 8/((x^2+4))

tangent of

\frac{8}{( x ^{2} + 4 )}

y = - \frac{16 a _{0}}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{24 a _{0}^{2} + 32}{( 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{8}{a _{0}^{2} + 4})

Find the slope of

y = \frac{8}{x ^{2} + 4} : \frac{d y}{d x} = - \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Compute the slope of

y = \frac{8}{x ^{2} + 4}

x = a_{0} : m = - \frac{16 a _{0}}{( a _{0}^{2} + 4 ) ^{2}}

Find the line with slope m=

- \frac{16 a _{0}}{( a _{0}^{2} + 4 ) ^{2}}

and passing through

(a_{0}, \frac{8}{a _{0}^{2} + 4}) : y = - \frac{16 a _{0}}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{24 a _{0}^{2} + 32}{( a _{0}^{2} + 4 ) ^{2}}

y = - \frac{16 a _{0}}{( a _{0}^{2} + 4 ) ^{2}} x + \frac{24 a _{0}^{2} + 32}{( a _{0}^{2} + 4 ) ^{2}}

\frac{d}{d x} (lo g_{2} ((x - 1)^{3}))

f (x) = x \cdot e^{- x + 1}

4 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{e ^{x}})

x \to 5 - lim (8 - x - x^{2})

What is the tangent of 8/((x^2+4)) ?
The tangent of 8/((x^2+4)) is y=-(16a_{0})/((a_{0)^2+4)^2}x+(24a_{0}^2+32)/((a_{0)^2+4)^2}