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

解

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

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

解答ステップ

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

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

\frac{\partial}{\partial y} (ln (x y)) = \frac{1}{y}

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

簡素化

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

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