What is the tangent of y=4x^3-4x(-1) ?

The tangent of y=4x^3-4x(-1) is y=(12a_{0}^2+4)x-8a_{0}^3

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tangent of y=4x^3-4x(-1)

tangent of

y = 4 x^{3} - 4 x (- 1)

y = (12 a_{0}^{2} + 4) x - 8 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 4 x^{3} - 4 x (- 1) : \frac{d y}{d x} = 12 x^{2} + 4

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 = 12 a_{0}^{2} + 4

Find the line with slope m=

12 a_{0}^{2} + 4

and passing through

(a_{0}, 4 a_{0}^{3} + 4 a_{0}) : y = (12 a_{0}^{2} + 4) x - 8 a_{0}^{3}

y = (12 a_{0}^{2} + 4) x - 8 a_{0}^{3}

ln (arccsc (x))

\frac{\partial}{\partial y} (x^{3} y^{5})

y^{^{'}} = \frac{x - 2}{y - 4}

f (x) = x^{2} + 9, at x = - 4

\int x os (x) d x

What is the tangent of y=4x^3-4x(-1) ?
The tangent of y=4x^3-4x(-1) is y=(12a_{0}^2+4)x-8a_{0}^3