What is the tangent of y=(x^3-9x)^{12},(-3,0) ?

The tangent of y=(x^3-9x)^{12},(-3,0) is y=0

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tangent of y=(x^3-9x)^{12},(-3,0)

tangent of

y = (x^{3} - 9 x)^{12}, (- 3, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 9 x)^{12} : \frac{d y}{d x} = 12 (x^{3} - 9 x)^{11} (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

\int (4 sin (x) + 2 cos (x)) d x

x^{2} (1 - 5 x)

\frac{\partial}{\partial u} (a sin (u) cos (v))

f (x) = \frac{216}{x ^{2} + 36}

\frac{d x}{d t} = \frac{4}{x e ^{t + 3 x}}

What is the tangent of y=(x^3-9x)^{12},(-3,0) ?
The tangent of y=(x^3-9x)^{12},(-3,0) is y=0