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}

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\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}}

\frac{d}{d q} (5 - ∣ 2 q - 3 ∣)

(x + 5)^{- 1}

(e^{x})^{\frac{1}{3}}

\frac{1}{( s ^{2} + 2 )}

\frac{d}{d x} (lo g_{10} (1 + x^{2} y^{2}))

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}