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人気のある微分積分 >

derivative y=ln(sqrt(2x+1))

解

導関数

y = ln (2 x + 1)

解

\frac{1}{2 x + 1}

解答ステップ

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

連鎖法則を適用:

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

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

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

= \frac{1}{2 x + 1} \cdot \frac{1}{2 x + 1}

簡素化

\frac{1}{2 x + 1} \cdot \frac{1}{2 x + 1} : \frac{1}{2 x + 1}

= \frac{1}{2 x + 1}

グラフ

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