What is the tangent of y=2x^3-2x(-1) ?

The tangent of y=2x^3-2x(-1) is y=(6a_{0}^2+2)x-4a_{0}^3

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tangent of y=2x^3-2x(-1)

tangent of

y = 2 x^{3} - 2 x (- 1)

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 (- 1) : \frac{d y}{d x} = 6 x^{2} + 2

Compute the slope of

y = 2 x^{3} - 2 x (- 1)

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}

\frac{d}{d x} (2 cos (- 2 x))

\int \frac{34 x ^{2}}{( 2 x + 1 ) ( x ^{2} + 4 )} d x

y^{^{'}} = - 3 x y^{3}

\int (6 x y) d x

e^{3 x}, 2

What is the tangent of y=2x^3-2x(-1) ?
The tangent of y=2x^3-2x(-1) is y=(6a_{0}^2+2)x-4a_{0}^3