What is the tangent of y=x^3-9x ?

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

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tangent of y=x^3-9x

tangent of

y = x^{3} - 9 x

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

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} - 9

Find the line with slope m=

3 a_{0}^{2} - 9

and passing through

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

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

\int - \frac{1}{x} d x

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

\frac{\partial}{\partial x} (x^{2} + y^{3})

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

\frac{d}{d x} (y - \frac{1}{( x + 3 )})

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