derivative of log_{10}(|1-2x^2|)

解

\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

- \frac{4 x}{ln ( 10 ) ( - 2 x ^{2} + 1 )}

解答ステップ

\frac{d}{d x} (lo g_{10} (1 - 2 x^{2}))

対数の規則を適用する:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln 1 - 2 x ^{2}}{ln ( 10 )})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln 1 - 2 x^{2})

連鎖法則を適用:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{∣ 1 - 2 x ^{2} ∣} (- \frac{4 x ( 1 - 2 x ^{2} )}{∣ 1 - 2 x ^{2} ∣})

簡素化

\frac{1}{ln ( 10 )} \cdot \frac{1}{∣ 1 - 2 x ^{2} ∣} (- \frac{4 x ( 1 - 2 x ^{2} )}{∣ 1 - 2 x ^{2} ∣}) : - \frac{4 x}{ln ( 10 ) ( - 2 x ^{2} + 1 )}

= - \frac{4 x}{ln ( 10 ) ( - 2 x ^{2} + 1 )}