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人気のある微分積分 >

d/(d{r)}((2k)/({r)}+2pi{r}^2)

解

\frac{d}{d r} (\frac{2 k}{r} + 2 π r^{2})

解

- \frac{2 k}{r ^{2}} + 4 π r

解答ステップ

\frac{d}{d r} (\frac{2 k}{r} + 2 π r^{2})

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d r} (\frac{2 k}{r}) + \frac{d}{d r} (2 π r^{2})

\frac{d}{d r} (\frac{2 k}{r}) = - \frac{2 k}{r ^{2}}

\frac{d}{d r} (2 π r^{2}) = 4 π r

= - \frac{2 k}{r ^{2}} + 4 π r

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