(\partial)/(\partial a)((h)(b+cos(ab)))

解

\frac{\partial}{\partial a} ((h) (b + cos (ab)))

- hb sin (ba)

解答ステップ

\frac{\partial}{\partial a} ((h) (b + cos (ab)))

定数を除去:

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

= h \frac{\partial}{\partial a} (b + cos (ab))

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= h (\frac{\partial}{\partial a} (b) + \frac{\partial}{\partial a} (cos (ab)))

\frac{\partial}{\partial a} (b) = 0

\frac{\partial}{\partial a} (cos (ab)) = - sin (ab) b

= h (0 - sin (ab) b)

簡素化

= - hb sin (ba)