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}

Compute the slope of

f (x) = (ln (x))^{2}

x = 11 : 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)

n = 1 \sum \infty 2 n

\frac{d y}{d x} - (2 + y) x^{2} = 0

\frac{\partial}{\partial x} (tan (x^{2} + y))

y = sin (x), y = cos (x), x = \frac{5 π}{4}, x = 2 π

y = \frac{2 x}{x + 2}, (2, 1)

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)