(\partial)/(\partial t)(ln(t^2+1))

解

\frac{\partial}{\partial t} (ln (t^{2} + 1))

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

解答ステップ

\frac{\partial}{\partial t} (ln (t^{2} + 1))

連鎖法則を適用:

\frac{1}{t ^{2} + 1} \frac{\partial}{\partial t} (t^{2} + 1)

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

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

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

簡素化

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

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