What is the tangent of y=(x^3-9x)^8,(-3,0) ?

The tangent of y=(x^3-9x)^8,(-3,0) is y=0

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tangent of y=(x^3-9x)^8,(-3,0)

tangent of

y = (x^{3} - 9 x)^{8}, (- 3, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 9 x)^{8} : \frac{d y}{d x} = 8 (x^{3} - 9 x)^{7} (3 x^{2} - 9)

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 = 0

Find the line with slope m=

0

and passing through

(- 3, 0) : y = 0

y = 0

f (x) = \frac{1}{x - 3}

d er i v a t i v e x (2 x - 5)^{3}

\frac{\partial}{\partial x} (arcsec (\frac{y}{x}))

\int \frac{x ^{2}}{x - 1} d x

\int (x^{2} - 6)^{4} (2 x) d x

What is the tangent of y=(x^3-9x)^8,(-3,0) ?
The tangent of y=(x^3-9x)^8,(-3,0) is y=0