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

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

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tangent of y=x^3-x

tangent of

y = 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}

\frac{d}{d x} (\frac{2 ( x + 8 ) ( x - 6 )}{x ^{2} - 7 ^{2}})

\frac{d}{d x} ((\frac{2 x + 4}{3 x - 1})^{4})

\int \frac{5 sin ( x )}{x} d x

y^{^{'}} - \frac{2}{t} y = t^{3}

\int 20 + 75 e^{- \frac{t}{50}} d t

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