# d/dr

#### schinb65

##### New member
I am having a little trouble remembering the rules with derivatives.

$\frac{a}{1-r}$ I know that it should be (derivative of the top*bottom - top*derivative bottom) / (bottom squared).

$\frac{d}{dr}\frac{a}{1-r}$ I tried this got the answer wrong, and looked up how to do this and they showed:

$a \frac{d}{dr}\frac{1}{1-r}$ Why was the a pulled out and the derivative not taken on it? Is it because we are taking the derivative with respect to r? since a is not r we do nothing with it?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Since $a$ is not a function of $r$ treat it as constant .

so $$\displaystyle \frac{d}{dr} \left(\frac{a}{1-r} \right) = a\frac{d}{dr} \left(\frac{1}{1-r} \right)$$

#### MarkFL

Staff member
I am having a little trouble remembering the rules with derivatives.

$\frac{a}{1-r}$ I know that it should be (derivative of the top*bottom - top*derivative bottom) / (bottom squared).

$\frac{d}{dr}\frac{a}{1-r}$ I tried this got the answer wrong, and looked up how to do this and they showed:

$a \frac{d}{dr}\frac{1}{1-r}$ Why was the a pulled out and the derivative not taken on it? Is it because we are taking the derivative with respect to r? since a is not r we do nothing with it?
In the case of the function you are given, with $a$ a constant, you could write:

$$\displaystyle f(r)=\frac{a}{1-r}=a(1-r)^{-1}$$

and simply apply the power/chain rules.

You are allowed to factor constants out of a function before differentiating:

$$\displaystyle \frac{d}{dx}\left(k\cdot f(x) \right)=k\frac{d}{dx}\left(f(x) \right)$$ where $k$ is a constant. If $k$ depends on $x$, then the product rule should be used.

#### Fantini

$$\frac{d}{dr} \frac{a}{1-r} = \frac{ \left( \frac{d}{dr} (a) \right) (1-r) - (a) \left( \frac{d}{dr} (1-r) \right)}{(1-r)^2} = \frac{0 - a(-1)}{(1-r)^2} = \frac{a}{(1-r)^2}.$$
We make use that $a$ is a constant when we say $d/dr (a) = 0$. So, whichever way you choose, you have to note that $a$ is constant.