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tangent 4x^2=ln(x-y)+3,\at x=0

解答

切线

4 x^{2} = ln (x - y) + 3, at x = 0

解答

y = x - \frac{1}{e ^{3}}

求解步骤

找到切点:

(0, - \frac{1}{e ^{3}})

计算

4 x^{2} = ln (x - y) + 3

的斜率:

\frac{d y}{d x} = - e^{4 x^{2} - 3} \cdot 8 x + 1

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 = 1

找到斜率 m=

1

且穿过

(0, - \frac{1}{e ^{3}})

的直线:

y = x - \frac{1}{e ^{3}}

y = x - \frac{1}{e ^{3}}

作图

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