derivative of x|x|^{|x|-1}(ln|x|+1)

解

\frac{d}{d x} (x ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1))

x^{2} ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x^{2} ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + ∣ x ∣^{∣ x ∣ - 1} + ln ∣ x ∣ ∣ x ∣^{∣ x ∣} + ∣ x ∣^{∣ x ∣}

解答ステップ

\frac{d}{d x} (x ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1)

= \frac{d x}{d x} ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1) + \frac{d}{d x} (∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1)) x

\frac{d x}{d x} = 1

\frac{d}{d x} (∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1)) = x ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + \frac{ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣} - ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣ - 1} + ∣ x ∣ ^{∣ x ∣}}{x}

= 1 \cdot ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1) + (x ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + \frac{ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣} - ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣ - 1} + ∣ x ∣ ^{∣ x ∣}}{x}) x

簡素化

1 \cdot ∣ x ∣^{∣ x ∣ - 1} (ln ∣ x ∣ + 1) + (x ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + \frac{ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣} - ln ∣ x ∣ ∣ x ∣ ^{∣ x ∣ - 1} + ∣ x ∣ ^{∣ x ∣}}{x}) x : x^{2} ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x^{2} ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + ∣ x ∣^{∣ x ∣ - 1} + ln ∣ x ∣ ∣ x ∣^{∣ x ∣} + ∣ x ∣^{∣ x ∣}

= x^{2} ln ∣ x ∣^{2} ∣ x ∣^{∣ x ∣ - 2} + x^{2} ln ∣ x ∣ ∣ x ∣^{∣ x ∣ - 2} + ∣ x ∣^{∣ x ∣ - 1} + ln ∣ x ∣ ∣ x ∣^{∣ x ∣} + ∣ x ∣^{∣ x ∣}