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

解

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

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

解答ステップ

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

x

を定数として扱う

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + y^{2}, g = ln (x^{2} + y^{2})

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

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

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

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

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

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