What is the normal of x/(1+x^2),\at x=2 ?

The normal of x/(1+x^2),\at x=2 is y= 25/3 x-244/15

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normal of x/(1+x^2),\at x=2

normal of

\frac{x}{1 + x ^{2}}, at x = 2

y = \frac{25}{3} x - \frac{244}{15}

Solution steps

Find the tangent point:

(2, \frac{2}{5})

Find the slope of

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

Compute the slope of

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

x = 2 : m = - \frac{3}{25}

Compute the slope of the perpendicular line:

m_{p} = \frac{25}{3}

Find the line with slope m=

\frac{25}{3}

and passing through

(2, \frac{2}{5}) : y = \frac{25}{3} x - \frac{244}{15}

y = \frac{25}{3} x - \frac{244}{15}

\frac{d}{d x} (arctan (e^{- 2 x}))

6 x - 2, x^{2} + 2 x + 2, 1

(- 1, 3), (- 6, - 2)

\int (2 + cos (2 x))^{2} d x

\int_{- 1}^{2} (2 - x^{2}) - (- x) d x

What is the normal of x/(1+x^2),\at x=2 ?
The normal of x/(1+x^2),\at x=2 is y= 25/3 x-244/15