What is the tangent of f(x)=(ln(x))^3,\at x=11 ?

The tangent of f(x)=(ln(x))^3,\at x=11 is y=(3ln^2(11))/(11)x+ln^3(11)-3ln^2(11)

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tangent of f(x)=(ln(x))^3,\at x=11

tangent of

f (x) = (ln (x))^{3}, at x = 11

y = \frac{3 ln ^{2} ( 11 )}{11} x + ln^{3} (11) - 3 ln^{2} (11)

Solution steps

Find the tangent point:

(11, ln^{3} (11))

Find the slope of

f (x) = (ln (x))^{3} : \frac{df ( x )}{d x} = \frac{3 ln ^{2} ( x )}{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 = \frac{3 ln ^{2} ( 11 )}{11}

Find the line with slope m=

\frac{3 ln ^{2} ( 11 )}{11}

and passing through

(11, ln^{3} (11)) : y = \frac{3 ln ^{2} ( 11 )}{11} x + ln^{3} (11) - 3 ln^{2} (11)

y = \frac{3 ln ^{2} ( 11 )}{11} x + ln^{3} (11) - 3 ln^{2} (11)

What is the tangent of f(x)=(ln(x))^3,\at x=11 ?
The tangent of f(x)=(ln(x))^3,\at x=11 is y=(3ln^2(11))/(11)x+ln^3(11)-3ln^2(11)