What is the tangent of x^3-x ?

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

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tangent of x^3-x

tangent of

x^{3} - x

y = (3 a_{0}^{2} - 1) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{3} - x : \frac{d y}{d x} = 3 x^{2} - 1

Compute the slope of

y = x^{3} - x

x = a_{0} : m = 3 a_{0}^{2} - 1

Find the line with slope m=

3 a_{0}^{2} - 1

and passing through

(a_{0}, a_{0}^{3} - a_{0}) : y = (3 a_{0}^{2} - 1) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 1) x - 2 a_{0}^{3}

\int (227 - x) - (23 + 0.7 x) d x

\int (2 x^{2} - 3)^{5} d x

\int e^{7 x} cos (4 x) d x

y^{^{'}} = (x + y - 3)^{2}

\int - x tan (x) d x