(\partial)/(\partial x)(2x^2+3y^2-20)

解

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - 20)

4 x

解答ステップ

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - 20)

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} (2 x^{2}) + \frac{\partial}{\partial x} (3 y^{2}) - \frac{\partial}{\partial x} (20)

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (20) = 0

= 4 x + 0 - 0

簡素化

= 4 x