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

解

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

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

解答ステップ

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

y

を定数として扱う

連鎖法則を適用:

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

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

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

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

簡素化

\frac{1}{2 ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} )} \cdot \frac{2 x}{x ^{2} + y ^{2}} : \frac{x}{ln ^{\frac{1}{2}} ( x ^{2} + y ^{2} ) ( x ^{2} + y ^{2} )}

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