(\partial)/(\partial y)(sqrt(2x^2-4y^3))

解

\frac{\partial}{\partial y} (2 x^{2} - 4 y^{3})

- \frac{6 y ^{2}}{2 x ^{2} - 4 y ^{3}}

解答ステップ

\frac{\partial}{\partial y} (2 x^{2} - 4 y^{3})

x

を定数として扱う

連鎖法則を適用:

\frac{1}{2 2 x ^{2} - 4 y ^{3}} \frac{\partial}{\partial y} (2 x^{2} - 4 y^{3})

= \frac{1}{2 2 x ^{2} - 4 y ^{3}} \frac{\partial}{\partial y} (2 x^{2} - 4 y^{3})

\frac{\partial}{\partial y} (2 x^{2} - 4 y^{3}) = - 12 y^{2}

= \frac{1}{2 2 x ^{2} - 4 y ^{3}} (- 12 y^{2})

簡素化

\frac{1}{2 2 x ^{2} - 4 y ^{3}} (- 12 y^{2}) : - \frac{6 y ^{2}}{2 x ^{2} - 4 y ^{3}}

= - \frac{6 y ^{2}}{2 x ^{2} - 4 y ^{3}}