What is the normal of y=(x+3)^3,\at x=-1/75 ?

The normal of y=(x+3)^3,\at x=-1/75 is y=-1875/50176 x+563938791749/21168000000

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normal of y=(x+3)^3,\at x=-1/75

normal of

y = (x + 3)^{3}, at x = - \frac{1}{75}

y = - \frac{1875}{50176} x + \frac{563938791749}{21168000000}

Solution steps

Find the tangent point:

(- \frac{1}{75}, \frac{11239424}{421875})

Find the slope of

y = (x + 3)^{3} : \frac{d y}{d x} = 3 (x + 3)^{2}

Compute the slope of

y = (x + 3)^{3}

x = - \frac{1}{75} : m = \frac{50176}{1875}

Compute the slope of the perpendicular line:

m_{p} = - \frac{1875}{50176}

Find the line with slope m=

- \frac{1875}{50176}

and passing through

(- \frac{1}{75}, \frac{11239424}{421875}) : y = - \frac{1875}{50176} x + \frac{563938791749}{21168000000}

y = - \frac{1875}{50176} x + \frac{563938791749}{21168000000}

\frac{d}{d x} (\frac{2}{( 2 - x ) ^{2}})

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

\frac{d}{d x} (arctan (5 x^{2} - 1))

\int x^{3} x^{2} + 9 d x

\frac{\partial}{\partial x} (sin (π (6 x - y)))

What is the normal of y=(x+3)^3,\at x=-1/75 ?
The normal of y=(x+3)^3,\at x=-1/75 is y=-1875/50176 x+563938791749/21168000000