What is the tangent of 5/(1+x^2) ?

The tangent of 5/(1+x^2) is y=-(10a_{0})/((1+a_{0)^2)^2}x+(15a_{0}^2+5)/((a_{0)^2+1)^2}

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tangent of 5/(1+x^2)

tangent of

\frac{5}{1 + x ^{2}}

y = - \frac{10 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( 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{5}{1 + a _{0}^{2}})

Find the slope of

y = \frac{5}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{10 x}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

y = \frac{5}{1 + x ^{2}}

x = a_{0} : m = - \frac{10 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

- \frac{10 a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{5}{1 + a _{0}^{2}}) : y = - \frac{10 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{10 a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( a _{0}^{2} + 1 ) ^{2}}

\frac{d}{d t} (\frac{sin ( t )}{1 + t ^{2}})

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

e^{- 2 x} sin (x), 0

\int \frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}})

What is the tangent of 5/(1+x^2) ?
The tangent of 5/(1+x^2) is y=-(10a_{0})/((1+a_{0)^2)^2}x+(15a_{0}^2+5)/((a_{0)^2+1)^2}