What is the tangent of x^3-3x-4 ?

The tangent of x^3-3x-4 is y=(3a_{0}^2-3)x-2a_{0}^3-4

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tangent of x^3-3x-4

tangent of

x^{3} - 3 x - 4

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} - 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0} - 4)

Find the slope of

y = x^{3} - 3 x - 4 : \frac{d y}{d x} = 3 x^{2} - 3

Compute the slope of

y = x^{3} - 3 x - 4

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

Find the line with slope m=

3 a_{0}^{2} - 3

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0} - 4) : y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} - 4

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} - 4

\frac{d}{d y} (\frac{2}{3} (y - 1)^{\frac{3}{2}})

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

x y^{^{'}} + 2 y \cdot ln (x) = 3

\int_{0}^{1} sin (nπ x) d x

\frac{3}{s} + \frac{24}{s ^{4}}

What is the tangent of x^3-3x-4 ?
The tangent of x^3-3x-4 is y=(3a_{0}^2-3)x-2a_{0}^3-4