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

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

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tangent of y=x^3-3x-4

tangent of

y = 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 y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}, y (1) = 9

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

2 t \cdot cos (t)

\int 3 t cos (2 t) d t

x \to 8 lim (\frac{∣ 8 - x ∣}{8 - x})

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