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)

Compute the slope of

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

(3, 0) : m = 0

Find the line with slope m=

0

and passing through

(3, 0) : y = 0

y = 0

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = x^{2} + 2

\frac{d}{d x} (\frac{0.9 e ^{x}}{tan ( x )})

(x^{\frac{1}{3}})^{^{'}}

3 t^{3} - \frac{2}{t ^{2}}

f (x) = x^{2} - 6

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