What is the normal of y=(sqrt(x))/(x+9),(1,0.1) ?

The normal of y=(sqrt(x))/(x+9),(1,0.1) is y=-25x+25.1

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normal of y=(sqrt(x))/(x+9),(1,0.1)

normal of

y = \frac{x}{x + 9}, (1, 0.1)

y = - 25 x + 25.1

Solution steps

Find the slope of

y = \frac{x}{x + 9} : \frac{d y}{d x} = \frac{- x + 9}{2 x ( x + 9 ) ^{2}}

Compute the slope of

y = \frac{x}{x + 9}

(1, 0.1) : m = \frac{1}{25}

Compute the slope of the perpendicular line:

m_{p} = - 25

Find the line with slope m=

- 25

and passing through

(1, 0.1) : y = - 25 x + 25.1

y = - 25 x + 25.1

x \to \infty lim (- 5 x^{2})

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

x^{^{'}} = 0.1078 \cdot (1 - \frac{x}{105900}) x - x

n = 1 \sum \infty \frac{15}{n}

\int_{0}^{1} x e^{2 x} d x

What is the normal of y=(sqrt(x))/(x+9),(1,0.1) ?
The normal of y=(sqrt(x))/(x+9),(1,0.1) is y=-25x+25.1