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tangent f(x)=ln(4-x^2+2x^4),\at x=1

解答

切线

f (x) = ln (4 - x^{2} + 2 x^{4}), at x = 1

解答

y = \frac{6}{5} x + ln (5) - \frac{6}{5}

求解步骤

找到切点:

(1, ln (5))

计算

f (x) = ln (4 - x^{2} + 2 x^{4})

的斜率:

\frac{df ( x )}{d x} = \frac{8 x ^{3} - 2 x}{4 - x ^{2} + 2 x ^{4}}

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{6}{5}

找到斜率 m=

\frac{6}{5}

且穿过

(1, ln (5))

的直线:

y = \frac{6}{5} x + ln (5) - \frac{6}{5}

y = \frac{6}{5} x + ln (5) - \frac{6}{5}

作图

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