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

The tangent of 2/(1+x^2) is y=-(4a_{0})/((1+a_{0)^2)^2}x+(6a_{0}^2+2)/((a_{0)^2+1)^2}

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tangent of 2/(1+x^2)

tangent of

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

y = - \frac{4 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{6 a _{0}^{2} + 2}{( 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{2}{1 + a _{0}^{2}})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = - \frac{4 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{2}{1 + a _{0}^{2}}) : y = - \frac{4 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{6 a _{0}^{2} + 2}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{4 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{6 a _{0}^{2} + 2}{( a _{0}^{2} + 1 ) ^{2}}

y^{^{'}} + y = e^{3 x}

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

\int \frac{1}{csc ^{4} ( 3 x ) sec ^{2} ( 3 x )} d x

\frac{\partial}{\partial y} (3 x y e^{- x^{2} - y^{2}})

\int sin (5 x) sin (2 x) d x

What is the tangent of 2/(1+x^2) ?
The tangent of 2/(1+x^2) is y=-(4a_{0})/((1+a_{0)^2)^2}x+(6a_{0}^2+2)/((a_{0)^2+1)^2}