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

Lösung

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

12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)

Schritte zur Lösung

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

Behandle

x

als Konstante

Entferne die Konstante:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

2 \cdot 2 sec (2 x + 3 y) sec (2 x + 3 y) tan (2 x + 3 y) \cdot 3 : 12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)

= 12 sec^{2} (2 x + 3 y) tan (2 x + 3 y)