What is the tangent of 4x^3-4x ?

The tangent of 4x^3-4x is y=(12a_{0}^2-4)x-8a_{0}^3

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tangent of 4x^3-4x

tangent of

4 x^{3} - 4 x

y = (12 a_{0}^{2} - 4) x - 8 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 4 a_{0})

Find the slope of

y = 4 x^{3} - 4 x : \frac{d y}{d x} = 12 x^{2} - 4

Compute the slope of

y = 4 x^{3} - 4 x

x = a_{0} : m = 12 a_{0}^{2} - 4

Find the line with slope m=

12 a_{0}^{2} - 4

and passing through

(a_{0}, 4 a_{0}^{3} - 4 a_{0}) : y = (12 a_{0}^{2} - 4) x - 8 a_{0}^{3}

y = (12 a_{0}^{2} - 4) x - 8 a_{0}^{3}

\frac{d}{d x} (x^{2} - 8 ln (x))

\int e^{x} d x

x \to 0 - lim (\frac{x ^{2}}{x})

y = (4 x - 1)^{3}

y = 7 x - x^{2}, at x = 1

What is the tangent of 4x^3-4x ?
The tangent of 4x^3-4x is y=(12a_{0}^2-4)x-8a_{0}^3