What is the tangent of y=(6x^2-6x+3)(1+2x),\at x=1 ?

The tangent of y=(6x^2-6x+3)(1+2x),\at x=1 is y=24x-15

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tangent of y=(6x^2-6x+3)(1+2x),\at x=1

tangent of

y = (6 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

y = 24 x - 15

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

y = (6 x^{2} - 6 x + 3) (1 + 2 x) : \frac{d y}{d x} = 36 x^{2} - 12 x

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 = 24

Find the line with slope m=

24

and passing through

(1, 9) : y = 24 x - 15

y = 24 x - 15

\int e^{y} y - e^{y} d y

\int \frac{1}{e ^{x} e ^{\frac{1}{x ^{2}}}} d x

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

(x - 9) y^{^{'}} + y = x^{2} - 10

\int \frac{1}{x ^{2} + x x} d x

What is the tangent of y=(6x^2-6x+3)(1+2x),\at x=1 ?
The tangent of y=(6x^2-6x+3)(1+2x),\at x=1 is y=24x-15