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

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

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

tangent of

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

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

Find the slope of

y = x^{4} - 4 x^{3} + 2 : \frac{d y}{d x} = 4 x^{3} - 12 x^{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} - 12 a_{0}^{2}

Find the line with slope m=

4 a_{0}^{3} - 12 a_{0}^{2}

and passing through

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

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

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