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

解答

切线

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

解答

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

求解步骤

找到切点:

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

计算

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

的斜率:

\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}

找到斜率 m=

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

且穿过

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

的直线:

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)

作图

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