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

The tangent of y=(x^2-x)ln(6x),\at x=2 is y=(3ln(12)+1)x-4ln(12)-2

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tangent of y=(x^2-x)ln(6x),\at x=2

tangent of

y = (x^{2} - x) ln (6 x), at x = 2

y = (3 ln (12) + 1) x - 4 ln (12) - 2

Solution steps

Find the tangent point:

(2, 2 ln (12))

Find the slope of

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

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 = 3 ln (12) + 1

Find the line with slope m=

3 ln (12) + 1

and passing through

(2, 2 ln (12)) : y = (3 ln (12) + 1) x - 4 ln (12) - 2

y = (3 ln (12) + 1) x - 4 ln (12) - 2

\int x x + 2 d x

\int \frac{1}{( 5 + x ^{2} )} d x

d er i v a t i v e f (x) = \frac{x ^{2} + 3}{x ^{2}}

d er i v a t i v e y = cos^{2} (3 x - 1)

x \to 0 - lim (\frac{8}{5 x ^{\frac{1}{2}}})

What is the tangent of y=(x^2-x)ln(6x),\at x=2 ?
The tangent of y=(x^2-x)ln(6x),\at x=2 is y=(3ln(12)+1)x-4ln(12)-2