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

The tangent of 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 4/(1+x^2)

tangent of

\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

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

Compute the slope of

y = \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}}

f (x) = cos (x) - x sin (x)

\frac{d}{d x} (e^{x} (x^{2} - 3 x + 1))

\int (x^{2} - 2 x)^{3} d x

f (t) = \frac{8}{t ^{2}}, at t = 1

\frac{d}{d x} ((2 x - 7)^{- 1} (x + 5))

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