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

The tangent of y=3x^3-3x is y=(9a_{0}^2-3)x-6a_{0}^3

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tangent of y=3x^3-3x

tangent of

y = 3 x^{3} - 3 x

y = (9 a_{0}^{2} - 3) x - 6 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 3 x^{3} - 3 x : \frac{d y}{d x} = 9 x^{2} - 3

Compute the slope of

y = 3 x^{3} - 3 x

x = a_{0} : m = 9 a_{0}^{2} - 3

Find the line with slope m=

9 a_{0}^{2} - 3

and passing through

(a_{0}, 3 a_{0}^{3} - 3 a_{0}) : y = (9 a_{0}^{2} - 3) x - 6 a_{0}^{3}

y = (9 a_{0}^{2} - 3) x - 6 a_{0}^{3}

\frac{d y}{d x} = 5 e^{x}

\frac{d}{d x} (ln (1 + cos (x)))

\int_{0}^{1} x arcsin (x^{2}) d x

\frac{d}{d x} (1200 x^{2} - 0.1 x)

\frac{d}{d x} (\frac{1}{x ^{3} + 1})

What is the tangent of y=3x^3-3x ?
The tangent of y=3x^3-3x is y=(9a_{0}^2-3)x-6a_{0}^3