What is the tangent of x^3-8x ?

The tangent of x^3-8x is y=(3a_{0}^2-8)x-2a_{0}^3

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tangent of x^3-8x

tangent of

x^{3} - 8 x

y = (3 a_{0}^{2} - 8) 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} - 8 a_{0})

Find the slope of

y = x^{3} - 8 x : \frac{d y}{d x} = 3 x^{2} - 8

Compute the slope of

y = x^{3} - 8 x

x = a_{0} : m = 3 a_{0}^{2} - 8

Find the line with slope m=

3 a_{0}^{2} - 8

and passing through

(a_{0}, a_{0}^{3} - 8 a_{0}) : y = (3 a_{0}^{2} - 8) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 8) x - 2 a_{0}^{3}

\int 18 x^{\frac{1}{2}} d x

\frac{\partial}{\partial x} (- \frac{1}{( x - y ) ^{2}})

f (y) = cos (y)

\int_{0}^{5} (\frac{x}{x + 4}) d x

y^{^{'}} + y + 1 = 0

What is the tangent of x^3-8x ?
The tangent of x^3-8x is y=(3a_{0}^2-8)x-2a_{0}^3