What is the normal of y= x/(1+x^2),(2, 2/5) ?

The normal of y= x/(1+x^2),(2, 2/5) is y= 25/3 x-244/15

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normal of y= x/(1+x^2),(2, 2/5)

normal of

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

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

Solution steps

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}}

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

n = 0 \sum \infty e^{- \frac{9 n}{2}}

\int u^{- 3} d u

\frac{d}{d t} (t^{\frac{3}{2}})

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

(x sinh (x) + cos (x))^{5}

What is the normal of y= x/(1+x^2),(2, 2/5) ?
The normal of y= x/(1+x^2),(2, 2/5) is y= 25/3 x-244/15