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

解

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

\frac{6 y}{2 x ^{3} + 3 y ^{2}}

解答ステップ

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

x

を定数として扱う

連鎖法則を適用:

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

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

\frac{\partial}{\partial y} (2 x^{3} + 3 y^{2}) = 6 y

= \frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y

簡素化

\frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y : \frac{6 y}{2 x ^{3} + 3 y ^{2}}

= \frac{6 y}{2 x ^{3} + 3 y ^{2}}