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)

Compute the slope of

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

(3, 0) : m = 0

Find the line with slope m=

0

and passing through

(3, 0) : y = 0

y = 0

\frac{\partial}{\partial x} (3 y cos (- 3 x))

\frac{d}{d x} (x^{2} - x - 12)

tan (7) x^{2}

\int_{0}^{1} 2 - x^{2} d x

y = \frac{x}{1 + x ^{2}}, (0, 0)

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