(\partial)/(\partial x)(e^{xy}cos(bx+c))

解

\frac{\partial}{\partial x} (e^{x y} cos (b x + c))

e^{x y} y cos (b x + c) - b e^{x y} sin (b x + c)

解答ステップ

\frac{\partial}{\partial x} (e^{x y} cos (b x + c))

b, c, y

を定数として扱う

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x y}, g = cos (b x + c)

= \frac{\partial}{\partial x} (e^{x y}) cos (b x + c) + \frac{\partial}{\partial x} (cos (b x + c)) e^{x y}

\frac{\partial}{\partial x} (e^{x y}) = e^{x y} y

\frac{\partial}{\partial x} (cos (b x + c)) = - sin (b x + c) b

= e^{x y} y cos (b x + c) + (- sin (b x + c) b) e^{x y}

簡素化

= e^{x y} y cos (b x + c) - b e^{x y} sin (b x + c)