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tangent f(x)=ln(6-x^2+2x^4),\at x=1

解答

切线

f (x) = ln (6 - x^{2} + 2 x^{4}), at x = 1

解答

y = \frac{6}{7} x + ln (7) - \frac{6}{7}

求解步骤

找到切点:

(1, ln (7))

计算

f (x) = ln (6 - x^{2} + 2 x^{4})

的斜率:

\frac{df ( x )}{d x} = \frac{8 x ^{3} - 2 x}{6 - 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}{7}

找到斜率 m=

\frac{6}{7}

且穿过

(1, ln (7))

的直线:

y = \frac{6}{7} x + ln (7) - \frac{6}{7}

y = \frac{6}{7} x + ln (7) - \frac{6}{7}

作图

流行的例子

(dy)/(dx)+(3/x)y=-x+4

\frac{d y}{d x} + (\frac{3}{x}) y = - x + 4

合并 1/3 tan^3(x)-tan(x)+x

j o in \frac{1}{3} tan^{3} (x) - tan (x) + x

limit as x approaches-8 of (x+2)/(x+8)

x \to - 8 lim (\frac{x + 2}{x + 8})

limit as x approaches 2 of 2x^2+7x-1

x \to 2 lim (2 x^{2} + 7 x - 1)

limit as x approaches-3+of 2x^2-6

x \to - 3 + lim (2 x^{2} - 6)