derivative of ln(2x^4-x^3+3x^2-3x)

解答

\frac{d}{d x} (ln (2 x^{4} - x^{3} + 3 x^{2} - 3 x))

\frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

求解步骤

\frac{d}{d x} (ln (2 x^{4} - x^{3} + 3 x^{2} - 3 x))

使用连锁律:

\frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} \frac{d}{d x} (2 x^{4} - x^{3} + 3 x^{2} - 3 x)

= \frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} \frac{d}{d x} (2 x^{4} - x^{3} + 3 x^{2} - 3 x)

\frac{d}{d x} (2 x^{4} - x^{3} + 3 x^{2} - 3 x) = 8 x^{3} - 3 x^{2} + 6 x - 3

= \frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} (8 x^{3} - 3 x^{2} + 6 x - 3)

化简

\frac{1}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x} (8 x^{3} - 3 x^{2} + 6 x - 3) : \frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

= \frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

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