What is the tangent of f(x)= 4/(1+x^2) ?

The tangent of f(x)= 4/(1+x^2) is y=-(8a_{0})/((1+a_{0)^2)^2}x+(12a_{0}^2+4)/((a_{0)^2+1)^2}

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tangent of f(x)= 4/(1+x^2)

tangent

y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4}{1 + a _{0}^{2}})

Find the slope of

f (x) = \frac{4}{1 + x ^{2}} : \frac{df ( x )}{d x} = - \frac{8 x}{( 1 + x ^{2} ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

- \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{4}{1 + a _{0}^{2}}) : y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{8 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{12 a _{0}^{2} + 4}{( a _{0}^{2} + 1 ) ^{2}}

What is the tangent of f(x)= 4/(1+x^2) ?
The tangent of f(x)= 4/(1+x^2) is y=-(8a_{0})/((1+a_{0)^2)^2}x+(12a_{0}^2+4)/((a_{0)^2+1)^2}