What is the tangent of f(x)= 9/(x^2+5) ?

The tangent of f(x)= 9/(x^2+5) is y=-(18a_{0})/((a_{0)^2+5)^2}x+(27a_{0}^2+45)/((a_{0)^2+5)^2}

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tangent of f(x)= 9/(x^2+5)

tangent of

f (x) = \frac{9}{x ^{2} + 5}

y = - \frac{18 a _{0}}{( a _{0}^{2} + 5 ) ^{2}} x + \frac{27 a _{0}^{2} + 45}{( a _{0}^{2} + 5 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{9}{a _{0}^{2} + 5})

Find the slope of

f (x) = \frac{9}{x ^{2} + 5} : \frac{df ( x )}{d x} = - \frac{18 x}{( x ^{2} + 5 ) ^{2}}

Compute the slope of

f (x) = \frac{9}{x ^{2} + 5}

x = a_{0} : m = - \frac{18 a _{0}}{( a _{0}^{2} + 5 ) ^{2}}

Find the line with slope m=

- \frac{18 a _{0}}{( a _{0}^{2} + 5 ) ^{2}}

and passing through

(a_{0}, \frac{9}{a _{0}^{2} + 5}) : y = - \frac{18 a _{0}}{( a _{0}^{2} + 5 ) ^{2}} x + \frac{27 a _{0}^{2} + 45}{( a _{0}^{2} + 5 ) ^{2}}

y = - \frac{18 a _{0}}{( a _{0}^{2} + 5 ) ^{2}} x + \frac{27 a _{0}^{2} + 45}{( a _{0}^{2} + 5 ) ^{2}}

cos^{2} (x), 0

sin (x) + \frac{4}{9} cot (x)

\int_{1}^{t} \frac{17}{x ^{3}} d x

3 - \frac{1}{2} x, 3 \leq x \leq 15

\int_{0}^{2} (x + 2 - x^{2}) d x

What is the tangent of f(x)= 9/(x^2+5) ?
The tangent of f(x)= 9/(x^2+5) is y=-(18a_{0})/((a_{0)^2+5)^2}x+(27a_{0}^2+45)/((a_{0)^2+5)^2}