en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

- \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

Popular Examples

y^{''}-4y^{''}+2y=0

y^{^{''}} - 4 y^{^{''}} + 2 y = 0

tangent of f(x)=x^2-4,\at x=1 tangent of

f (x) = x^{2} - 4, at x = 1

(y^2+9)dx-1/((2x+1)^2)dy=0

(y^{2} + 9) d x - \frac{1}{( 2 x + 1 ) ^{2}} d y = 0

derivative of ((x+3)/(x-3))^5 derivative of

(\frac{x + 3}{x - 3})^{5}

integral of 5xsqrt(1-x^2)

\int 5 x 1 - x^{2} d x

Frequently Asked Questions (FAQ)

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}