derivative sec^2(2x+5)*2

解

導関数

sec^{2} (2 x + 5) \cdot 2

8 sec^{2} (2 x + 5) tan (2 x + 5)

解答ステップ

\frac{d}{d x} (sec^{2} (2 x + 5) \cdot 2)

定数を除去:

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

= 2 \frac{d}{d x} (sec^{2} (2 x + 5))

連鎖法則を適用:

2 sec (2 x + 5) \frac{d}{d x} (sec (2 x + 5))

= 2 sec (2 x + 5) \frac{d}{d x} (sec (2 x + 5))

\frac{d}{d x} (sec (2 x + 5)) = sec (2 x + 5) tan (2 x + 5) \cdot 2

= 2 \cdot 2 sec (2 x + 5) sec (2 x + 5) tan (2 x + 5) \cdot 2

簡素化

2 \cdot 2 sec (2 x + 5) sec (2 x + 5) tan (2 x + 5) \cdot 2 : 8 sec^{2} (2 x + 5) tan (2 x + 5)

= 8 sec^{2} (2 x + 5) tan (2 x + 5)