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

The tangent of y=4x^3,(-1,-4) is y=12x+8

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

tangent of

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

y = 12 x + 8

Solution steps

Find the slope of

y = 4 x^{3} : \frac{d y}{d x} = 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 = 12

Find the line with slope m=

12

and passing through

(- 1, - 4) : y = 12 x + 8

y = 12 x + 8

x \to \frac{1}{2} lim (\frac{x + 1}{x + 5})

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

\frac{1}{1 - x}, 3

\frac{d}{d x} (\frac{5}{1 - x})

\int \frac{1}{sec ^{2} ( y )} d y

What is the tangent of y=4x^3,(-1,-4) ?
The tangent of y=4x^3,(-1,-4) is y=12x+8