derivative of ln(x^2+y^2+1)

解

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

\frac{2 x + 2 y \frac{d y}{d x}}{x ^{2} + y ^{2} + 1}

解答ステップ

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

連鎖法則を適用:

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

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

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

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

簡素化

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

= \frac{2 x + 2 y \frac{d y}{d x}}{x ^{2} + y ^{2} + 1}