derivative (tanh(4x+1))/5

解

導関数

\frac{tanh ( 4 x + 1 )}{5}

\frac{4}{5} sech^{2} (4 x + 1)

解答ステップ

\frac{d}{d x} (\frac{tanh ( 4 x + 1 )}{5})

定数を除去:

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

= \frac{1}{5} \frac{d}{d x} (tanh (4 x + 1))

連鎖法則を適用:

sech^{2} (4 x + 1) \frac{d}{d x} (4 x + 1)

= sech^{2} (4 x + 1) \frac{d}{d x} (4 x + 1)

\frac{d}{d x} (4 x + 1) = 4

= \frac{1}{5} sech^{2} (4 x + 1) \cdot 4

簡素化

\frac{1}{5} sech^{2} (4 x + 1) \cdot 4 : \frac{4}{5} sech^{2} (4 x + 1)

= \frac{4}{5} sech^{2} (4 x + 1)