What is the d/(d{r)}(1/(2{r)(a+b+c)}) ?

The d/(d{r)}(1/(2{r)(a+b+c)}) is -1/(2(a+b+c){r)^2}

What is the first d/(d{r)}(1/(2{r)(a+b+c)}) ?

The first d/(d{r)}(1/(2{r)(a+b+c)}) is -1/(2(a+b+c){r)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d r} (\frac{1}{2 r ( a + b + c )})

- \frac{1}{2 ( a + b + c ) r ^{2}}

Solution steps

\frac{d}{d r} (\frac{1}{2 r ( a + b + c )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2 ( a + b + c )} \frac{d}{d r} (\frac{1}{r})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{1}{2 ( a + b + c )} \frac{d}{d r} (r^{- 1})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{2 ( a + b + c )} (- 1 \cdot r^{- 1 - 1})

Simplify

\frac{1}{2 ( a + b + c )} (- 1 \cdot r^{- 1 - 1}) : - \frac{1}{2 ( a + b + c ) r ^{2}}

= - \frac{1}{2 ( a + b + c ) r ^{2}}

d er i v a t i v e (x + 5)^{- 1}

d er i v a t i v e (e^{x})^{\frac{1}{3}}

d er i v a t i v e \frac{1}{( s ^{2} + 2 )}

\frac{d}{d t} (t^{3} - 6 t^{2} + 9 t)

d er i v a t i v e \frac{( 1 + e ^{x} )}{( 1 - e ^{x} )}

What is the d/(d{r)}(1/(2{r)(a+b+c)}) ?
The d/(d{r)}(1/(2{r)(a+b+c)}) is -1/(2(a+b+c){r)^2}
What is the first d/(d{r)}(1/(2{r)(a+b+c)}) ?
The first d/(d{r)}(1/(2{r)(a+b+c)}) is -1/(2(a+b+c){r)^2}