(\partial)/(\partial y)(x^5+2^y+x^y)

解

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

x^{y} ln (x) + ln (2) \cdot 2^{y}

解答ステップ

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

x

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 y} (x^{5}) + \frac{\partial}{\partial y} (2^{y}) + \frac{\partial}{\partial y} (x^{y})

\frac{\partial}{\partial y} (x^{5}) = 0

\frac{\partial}{\partial y} (2^{y}) = 2^{y} ln (2)

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

= 0 + 2^{y} ln (2) + x^{y} ln (x)

簡素化

= x^{y} ln (x) + ln (2) \cdot 2^{y}