What is the tangent of f(x)=x^3-x ?

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

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tangent of f(x)=x^3-x

tangent of

f (x) = 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

f (x) = x^{3} - x : \frac{df ( x )}{d x} = 3 x^{2} - 1

Compute the slope of

f (x) = 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 4 t^{3} cos (t^{4} + 1) d t

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

5 - 2 x

\int x (x^{2} - 1)^{99} d x

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