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

解

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

\frac{4 x}{2 x ^{2} + y ^{2}}

解答ステップ

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

y

を定数として扱う

連鎖法則を適用:

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

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

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

= \frac{1}{2 x ^{2} + y ^{2}} \cdot 4 x

簡素化

\frac{1}{2 x ^{2} + y ^{2}} \cdot 4 x : \frac{4 x}{2 x ^{2} + y ^{2}}

= \frac{4 x}{2 x ^{2} + y ^{2}}