What is the normal of y=sqrt(x),\at x=4 ?

The normal of y=sqrt(x),\at x=4 is y=-4x+18

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normal of y=sqrt(x),\at x=4

normal of

y = x, at x = 4

y = - 4 x + 18

Solution steps

Find the tangent point:

(4, 2)

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

x = 4 : m = \frac{1}{4}

Compute the slope of the perpendicular line:

m_{p} = - 4

Find the line with slope m=

- 4

and passing through

(4, 2) : y = - 4 x + 18

y = - 4 x + 18

f (x) = 9 - x^{2}, at x = - 1

x^{- \frac{2}{3}}

\int \frac{7 x + 6}{x ^{2} ( x + 2 )} d x

y = 8 x^{2} + 8, (2, 40)

\frac{\partial}{\partial z} (x e^{y - z^{2}})

What is the normal of y=sqrt(x),\at x=4 ?
The normal of y=sqrt(x),\at x=4 is y=-4x+18