derivative arctan((1+t)/(1-t))

解

導関数

arctan (\frac{1 + t}{1 - t})

\frac{1}{t ^{2} + 1}

解答ステップ

\frac{d}{d t} (arctan (\frac{1 + t}{1 - t}))

連鎖法則を適用:

\frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \frac{d}{d t} (\frac{1 + t}{1 - t})

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

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

= \frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \cdot \frac{2}{( 1 - t ) ^{2}}

簡素化

\frac{1}{( \frac{1 + t}{1 - t} ) ^{2} + 1} \cdot \frac{2}{( 1 - t ) ^{2}} : \frac{1}{t ^{2} + 1}

= \frac{1}{t ^{2} + 1}