(\partial)/(\partial x)(x^2+y^2-ycos(x))

解

\frac{\partial}{\partial x} (x^{2} + y^{2} - y cos (x))

2 x + y sin (x)

解答ステップ

\frac{\partial}{\partial x} (x^{2} + y^{2} - y cos (x))

y

を定数として扱う

和/差の法則を適用:

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

= \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (y cos (x))

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (y cos (x)) = - y sin (x)

= 2 x + 0 - (- y sin (x))

簡素化

= 2 x + y sin (x)