(\partial)/(\partial y)(-2sec^2(3x-2y+1))

解

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

解答ステップ

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

x

を定数として扱う

定数を除去:

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

= - 2 \frac{\partial}{\partial y} (sec^{2} (3 x - 2 y + 1))

連鎖法則を適用:

2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

= 2 sec (3 x - 2 y + 1) \frac{\partial}{\partial y} (sec (3 x - 2 y + 1))

\frac{\partial}{\partial y} (sec (3 x - 2 y + 1)) = - 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= - 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1))

簡素化

- 2 \cdot 2 sec (3 x - 2 y + 1) (- 2 sec (3 x - 2 y + 1) tan (3 x - 2 y + 1)) : 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)

= 8 sec^{2} (3 x - 2 y + 1) tan (3 x - 2 y + 1)