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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

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

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 = ln^{2} (3)

Find the line with slope m=

ln^{2} (3)

and passing through

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

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

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

y^{^{''''}} + 3 y^{^{''}} - 4 y = 0

arctan (\frac{x - 1}{x + 1})

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

\frac{6 s + 7}{s ^{2} - s - 6}

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