What is the tangent of y=2x^3-2x ?

The tangent of y=2x^3-2x is y=(6a_{0}^2-2)x-4a_{0}^3

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tangent of y=2x^3-2x

tangent of

y = 2 x^{3} - 2 x

y = (6 a_{0}^{2} - 2) x - 4 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - 2 a_{0})

Find the slope of

y = 2 x^{3} - 2 x : \frac{d y}{d x} = 6 x^{2} - 2

Compute the slope of

y = 2 x^{3} - 2 x

x = a_{0} : m = 6 a_{0}^{2} - 2

Find the line with slope m=

6 a_{0}^{2} - 2

and passing through

(a_{0}, 2 a_{0}^{3} - 2 a_{0}) : y = (6 a_{0}^{2} - 2) x - 4 a_{0}^{3}

y = (6 a_{0}^{2} - 2) x - 4 a_{0}^{3}

(cos^{2} (x))^{^{'}}

(- 6, - 4), (- 4, 0)

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

\frac{\partial}{\partial x} (x ln (x))

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

What is the tangent of y=2x^3-2x ?
The tangent of y=2x^3-2x is y=(6a_{0}^2-2)x-4a_{0}^3