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 of

f (x) = \frac{4}{1 + x ^{2}}

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}}

Compute the slope of

f (x) = \frac{4}{1 + x ^{2}}

x = a_{0} : 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}}

x y^{2} \frac{d y}{d x} = y^{3} - x^{3}

cos (- 2 t)

\int 23 x d x

y^{^{'}} = \frac{y}{t}

y^{^{''}} + 4 y = 16 sin (2 t)

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}