What is the tangent of y=x^4+3x^3-2x-2 ?

The tangent of y=x^4+3x^3-2x-2 is y=(4a_{0}^3+9a_{0}^2-2)x-3a_{0}^4-6a_{0}^3-2

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tangent of y=x^4+3x^3-2x-2

tangent of

y = x^{4} + 3 x^{3} - 2 x - 2

y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} + 3 a_{0}^{3} - 2 a_{0} - 2)

Find the slope of

y = x^{4} + 3 x^{3} - 2 x - 2 : \frac{d y}{d x} = 4 x^{3} + 9 x^{2} - 2

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 = 4 a_{0}^{3} + 9 a_{0}^{2} - 2

Find the line with slope m=

4 a_{0}^{3} + 9 a_{0}^{2} - 2

and passing through

(a_{0}, a_{0}^{4} + 3 a_{0}^{3} - 2 a_{0} - 2) : y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

lo g_{5} (x^{2} - 1)

x \to 1 lim (- x^{2} - 1)

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

\int 20 x e^{8 x} d x

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t

What is the tangent of y=x^4+3x^3-2x-2 ?
The tangent of y=x^4+3x^3-2x-2 is y=(4a_{0}^3+9a_{0}^2-2)x-3a_{0}^4-6a_{0}^3-2