What is the normal of y=(sqrt(x))/(x+6),(4,0.2) ?

The normal of y=(sqrt(x))/(x+6),(4,0.2) is y=-200x+800.2

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normal of y=(sqrt(x))/(x+6),(4,0.2)

normal of

y = \frac{x}{x + 6}, (4, 0.2)

y = - 200 x + 800.2

Solution steps

Find the slope of

y = \frac{x}{x + 6} : \frac{d y}{d x} = \frac{- x + 6}{2 x ( x + 6 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{200}

Compute the slope of the perpendicular line:

m_{p} = - 200

Find the line with slope m=

- 200

and passing through

(4, 0.2) : y = - 200 x + 800.2

y = - 200 x + 800.2

\int_{1}^{e} ln (12 x) d x

y = x^{2}, y = x^{3}

x y^{^{'}} + y = - 8 x y^{2}

f (x) = x^{2} sin (x)

\int x + 8 d x

What is the normal of y=(sqrt(x))/(x+6),(4,0.2) ?
The normal of y=(sqrt(x))/(x+6),(4,0.2) is y=-200x+800.2