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

The tangent of f(x)= 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 f(x)= 8/(x^2+4)

tangent of

f (x) = \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

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

Compute the slope of

f (x) = \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}}

\int 3 x^{3} + 4 x^{2} - 5 x d x

x \to 0 lim (\frac{2 ( y + x ) - 2 y}{x})

\int 6 x^{2} + 4 x - 1 d x

\int \frac{10 x ^{3} - 5 x}{x ^{4} - x ^{2} + 6} d x

\int 3 x \cdot e^{x} d x

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