What is the tangent of f(x)=(ln(x))^3,\at x=9 ?

The tangent of f(x)=(ln(x))^3,\at x=9 is y=(4ln^2(3))/3 x+8ln^3(3)-12ln^2(3)

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tangent of f(x)=(ln(x))^3,\at x=9

tangent of

f (x) = (ln (x))^{3}, at x = 9

y = \frac{4 ln ^{2} ( 3 )}{3} x + 8 ln^{3} (3) - 12 ln^{2} (3)

Solution steps

Find the tangent point:

(9, 8 ln^{3} (3))

Find the slope of

f (x) = (ln (x))^{3} : \frac{df ( x )}{d x} = \frac{3 ln ^{2} ( x )}{x}

Compute the slope of

f (x) = (ln (x))^{3}

x = 9 : m = \frac{4 ln ^{2} ( 3 )}{3}

Find the line with slope m=

\frac{4 ln ^{2} ( 3 )}{3}

and passing through

(9, 8 ln^{3} (3)) : y = \frac{4 ln ^{2} ( 3 )}{3} x + 8 ln^{3} (3) - 12 ln^{2} (3)

y = \frac{4 ln ^{2} ( 3 )}{3} x + 8 ln^{3} (3) - 12 ln^{2} (3)

x \to 0 lim (\frac{2 - 2 cos ( x )}{x})

\int_{a}^{b} x d x

y^{2} - 2 x = 9, x - y = 3 [- 4.5, 8]

\frac{d}{d x} (ln (arctan (3 x^{3})))

What is the tangent of f(x)=(ln(x))^3,\at x=9 ?
The tangent of f(x)=(ln(x))^3,\at x=9 is y=(4ln^2(3))/3 x+8ln^3(3)-12ln^2(3)