What is the d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) ?

The d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) is (400(-{r}+2))/((2+3{r))^3}

What is the first d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) ?

The first d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) is (400(-{r}+2))/((2+3{r))^3}

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\frac{d}{d r} (\frac{200 r ( r + 2 )}{( 2 + 3 r ) ^{2}})

\frac{400 ( - r + 2 )}{( 2 + 3 r ) ^{3}}

Solution steps

\frac{d}{d r} (\frac{200 r ( r + 2 )}{( 2 + 3 r ) ^{2}})

Take the constant out:

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

= 200 \frac{d}{d r} (\frac{r ( r + 2 )}{( 2 + 3 r ) ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 200 \cdot \frac{\frac{d}{d r} ( r ( r + 2 ) ) ( 2 + 3 r ) ^{2} - \frac{d}{d r} ( ( 2 + 3 r ) ^{2} ) r ( r + 2 )}{( ( 2 + 3 r ) ^{2} ) ^{2}}

\frac{d}{d r} (r (r + 2)) = 2 r + 2

\frac{d}{d r} ((2 + 3 r)^{2}) = 6 (2 + 3 r)

= 200 \cdot \frac{( 2 r + 2 ) ( 2 + 3 r ) ^{2} - 6 ( 2 + 3 r ) r ( r + 2 )}{( ( 2 + 3 r ) ^{2} ) ^{2}}

Simplify

200 \cdot \frac{( 2 r + 2 ) ( 2 + 3 r ) ^{2} - 6 ( 2 + 3 r ) r ( r + 2 )}{( ( 2 + 3 r ) ^{2} ) ^{2}} : \frac{400 ( - r + 2 )}{( 2 + 3 r ) ^{3}}

= \frac{400 ( - r + 2 )}{( 2 + 3 r ) ^{3}}

y = 9 - 10 x, at x = - 1

\frac{d}{d x} (2 x y^{3})

\frac{\partial}{\partial y} (e^{s i n (x + y)})

f (x) = 1 + 4 csc (x), at x = - \frac{π}{3}

n = 1 \sum \infty \frac{2}{3 n + 10}

What is the d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) ?
The d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) is (400(-{r}+2))/((2+3{r))^3}
What is the first d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) ?
The first d/(d{r)}((200{r}({r}+2))/((2+3{r))^2}) is (400(-{r}+2))/((2+3{r))^3}