What is the tangent of f(x)=(ln(x))^2,\at x=11 ?

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

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tangent of f(x)=(ln(x))^2,\at x=11

tangent of

f (x) = (ln (x))^{2}, at x = 11

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

Solution steps

Find the tangent point:

(11, ln^{2} (11))

Find the slope of

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

Find the line with slope m=

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

and passing through

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

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

\int_{0}^{2} x^{2} - x d x

x \to 5 lim (5 x - (x - 5))

n = 1 \sum \infty e^{- \frac{9 n}{2}}

\int - 2 e^{- 2 x} sin (- 2 y)

f (x) = x^{4} + \frac{2}{x}

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