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}

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}{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

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

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

\frac{d}{d x} (1 - 5 x)

\frac{d y}{d x} = \frac{1}{x ( ln ( 2 ) )}

\int \frac{( 2 x + 3 )}{x + 2} d x

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