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

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

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tangent of y=2x^3-2x^2-2x

tangent of

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

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 6 a_{0}^{2} - 4 a_{0} - 2

Find the line with slope m=

6 a_{0}^{2} - 4 a_{0} - 2

and passing through

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

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

\frac{d y}{d x} + \frac{y}{x} = x^{3}

\frac{d}{d x} (\frac{e ^{a x^{2} + b x + c}}{x})

e \to 1 lim (5 ln^{2} (e))

x \to + 0 lim ((1 - \frac{1}{x}) x)

\frac{\partial}{\partial y} (x + y + 1)

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