What is the normal of y= x/(sqrt(4+x^2)),(0,0) ?

The normal of y= x/(sqrt(4+x^2)),(0,0) is y=-2x

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normal of y= x/(sqrt(4+x^2)),(0,0)

normal of

y = \frac{x}{4 + x ^{2}}, (0, 0)

y = - 2 x

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{x}{4 + x ^{2}}

(0, 0) : m = \frac{1}{2}

Compute the slope of the perpendicular line:

m_{p} = - 2

Find the line with slope m=

- 2

and passing through

(0, 0) : y = - 2 x

y = - 2 x

\int \frac{x ^{2} - 8}{x ^{4}} d x

\int_{0}^{1} (x^{3} - 3 x^{2}) d x

\frac{d}{d t} (4 t)

\int_{0}^{\infty} e^{- k x} d x

(2 x - 1) d x + (7 y + 4) d y = 0

What is the normal of y= x/(sqrt(4+x^2)),(0,0) ?
The normal of y= x/(sqrt(4+x^2)),(0,0) is y=-2x