What is the d/(d{r)}((2k)/({r)}+2pi{r}^2) ?

The d/(d{r)}((2k)/({r)}+2pi{r}^2) is -(2k)/({r)^2}+4pi{r}

What is the first d/(d{r)}((2k)/({r)}+2pi{r}^2) ?

The first d/(d{r)}((2k)/({r)}+2pi{r}^2) is -(2k)/({r)^2}+4pi{r}

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

- \frac{2 k}{r ^{2}} + 4 π r

Solution steps

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

Apply the Sum/Difference Rule:

(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

What is the d/(d{r)}((2k)/({r)}+2pi{r}^2) ?
The d/(d{r)}((2k)/({r)}+2pi{r}^2) is -(2k)/({r)^2}+4pi{r}
What is the first d/(d{r)}((2k)/({r)}+2pi{r}^2) ?
The first d/(d{r)}((2k)/({r)}+2pi{r}^2) is -(2k)/({r)^2}+4pi{r}