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

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

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

normal of

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

f (x) = \frac{25}{3} x - \frac{244}{15}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \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}) : f (x) = \frac{25}{3} x - \frac{244}{15}

f (x) = \frac{25}{3} x - \frac{244}{15}

5 x + 11

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

x \to 0 lim (\frac{( x + 64 ) ^{\frac{1}{3}} - 4}{x})

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

\int_{1}^{\infty} x^{- \frac{1}{5}} d x

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