derivative of log_{10}(x/(sqrt(x^2+1)))

解

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

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

解答ステップ

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

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

定数を除去:

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

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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