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

解

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

3 x^{2} + y x^{y - 1}

解答ステップ

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

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

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

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

\frac{\partial}{\partial x} (x^{y}) = y x^{y - 1}

= 3 x^{2} + 0 + y x^{y - 1}

簡素化

= 3 x^{2} + y x^{y - 1}