What is the tangent of y=(x^3-16x)^8,(-4,0) ?

The tangent of y=(x^3-16x)^8,(-4,0) is y=0

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tangent of y=(x^3-16x)^8,(-4,0)

tangent of

y = (x^{3} - 16 x)^{8}, (- 4, 0)

y = 0

Solution steps

Find the slope of

y = (x^{3} - 16 x)^{8} : \frac{d y}{d x} = 8 (x^{3} - 16 x)^{7} (3 x^{2} - 16)

Compute the slope of

y = (x^{3} - 16 x)^{8}

(- 4, 0) : m = 0

Find the line with slope m=

0

and passing through

(- 4, 0) : y = 0

y = 0

\frac{d y}{d x} = (x + y)

\frac{d}{d x} (x (ln (x^{2})))

\int sin^{6} (2 x) d x

x^{^{''}} + 3 x^{^{'}} + 4 x = 0, x (0) = 3, x^{^{'}} (0) = 1

\frac{d}{d x} (- \frac{e ^{- ∣ x ∣} x}{∣ x ∣})

What is the tangent of y=(x^3-16x)^8,(-4,0) ?
The tangent of y=(x^3-16x)^8,(-4,0) is y=0