What is the tangent of 2x^3-4x^2 ?

The tangent of 2x^3-4x^2 is y=(6a_{0}^2-8a_{0})x-4a_{0}^3+4a_{0}^2

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tangent of 2x^3-4x^2

tangent of

2 x^{3} - 4 x^{2}

y = (6 a_{0}^{2} - 8 a_{0}) x - 4 a_{0}^{3} + 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - 4 a_{0}^{2})

Find the slope of

y = 2 x^{3} - 4 x^{2} : \frac{d y}{d x} = 6 x^{2} - 8 x

Compute the slope of

y = 2 x^{3} - 4 x^{2}

x = a_{0} : m = 6 a_{0}^{2} - 8 a_{0}

Find the line with slope m=

6 a_{0}^{2} - 8 a_{0}

and passing through

(a_{0}, 2 a_{0}^{3} - 4 a_{0}^{2}) : y = (6 a_{0}^{2} - 8 a_{0}) x - 4 a_{0}^{3} + 4 a_{0}^{2}

y = (6 a_{0}^{2} - 8 a_{0}) x - 4 a_{0}^{3} + 4 a_{0}^{2}

\int cos (4 x) cos (8 x) d x

\frac{d}{d x} (\frac{6}{3 x ^{5}} + \frac{2}{5 x ^{2}} - \frac{7}{9 x})

(y^{2} - x^{2}) d x + (- 2 x y) d y = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = 8 e^{- 2 x}

\frac{d}{d x} (x e^{x} cos (x))

What is the tangent of 2x^3-4x^2 ?
The tangent of 2x^3-4x^2 is y=(6a_{0}^2-8a_{0})x-4a_{0}^3+4a_{0}^2