What is the tangent of 3/(1+x^2) ?

The tangent of 3/(1+x^2) is y=-(6a_{0})/((1+a_{0)^2)^2}x+(9a_{0}^2+3)/((a_{0)^2+1)^2}

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tangent of 3/(1+x^2)

tangent of

\frac{3}{1 + x ^{2}}

y = - \frac{6 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{9 a _{0}^{2} + 3}{( 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{3}{1 + a _{0}^{2}})

Find the slope of

y = \frac{3}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{6 x}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

y = \frac{3}{1 + x ^{2}}

x = a_{0} : m = - \frac{6 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

- \frac{6 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{3}{1 + a _{0}^{2}}) : y = - \frac{6 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{9 a _{0}^{2} + 3}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{6 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{9 a _{0}^{2} + 3}{( a _{0}^{2} + 1 ) ^{2}}

x y - π cos (y) = 3 π, (2, π)

\frac{\partial}{\partial t} (\frac{2 r + s}{t})

\frac{d}{d x} (- 12 sin (6 x))

y = (3 x + 4)^{2}

x \to 0 lim (\frac{x ^{2} + 5}{3 ^{x}})

What is the tangent of 3/(1+x^2) ?
The tangent of 3/(1+x^2) is y=-(6a_{0})/((1+a_{0)^2)^2}x+(9a_{0}^2+3)/((a_{0)^2+1)^2}