What is the tangent of x^3-5x ?

The tangent of x^3-5x is y=(3a_{0}^2-5)x-2a_{0}^3

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tangent of x^3-5x

tangent of

x^{3} - 5 x

y = (3 a_{0}^{2} - 5) 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} - 5 a_{0})

Find the slope of

y = x^{3} - 5 x : \frac{d y}{d x} = 3 x^{2} - 5

Compute the slope of

y = x^{3} - 5 x

x = a_{0} : m = 3 a_{0}^{2} - 5

Find the line with slope m=

3 a_{0}^{2} - 5

and passing through

(a_{0}, a_{0}^{3} - 5 a_{0}) : y = (3 a_{0}^{2} - 5) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 5) x - 2 a_{0}^{3}

\int_{0}^{π} cos (3 x) cos (3 x) d x

\int \frac{2}{t ^{2} - 1} d t

\int \frac{- 2}{x ^{2} - 25} d x

\int sin (x) 2 + 3 cos (x) d x

y = (3 x^{2} + 2) (2 x - 5)

What is the tangent of x^3-5x ?
The tangent of x^3-5x is y=(3a_{0}^2-5)x-2a_{0}^3