What is the tangent of y=(x^3-4x)^8,(2,0) ?

The tangent of y=(x^3-4x)^8,(2,0) is y=0

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tangent of y=(x^3-4x)^8,(2,0)

tangent of

y = (x^{3} - 4 x)^{8}, (2, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 4 x)^{8} : \frac{d y}{d x} = 8 (x^{3} - 4 x)^{7} (3 x^{2} - 4)

Compute the slope of

y = (x^{3} - 4 x)^{8}

(2, 0) : m = 0

Find the line with slope m=

0

and passing through

(2, 0) : y = 0

y = 0

\frac{d}{d t} (\frac{e ^{t b} - e ^{t a}}{t ( b - a )})

y^{^{''}} + 9 y = 12 sin (3 t)

x \to 2 lim (\frac{x ^{3}}{3} - 7 x^{2})

f (y) = y^{4}

\frac{d ^{2}}{d x ^{2}} (sec (\frac{π}{4}))

What is the tangent of y=(x^3-4x)^8,(2,0) ?
The tangent of y=(x^3-4x)^8,(2,0) is y=0