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}}

f (x) = \frac{20 x}{x ^{2} - 5}

x \to - 6 + lim (\frac{1}{( x + 6 ) ^{6}})

\int_{0}^{3} - 6 x^{2} d x

f (x) = 2 x^{3} - 5 x + 4, at x = - 1

\frac{1}{3 x ^{3}} + \frac{x ^{7}}{10}

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}