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d/dr

schinb65

New member
Jan 1, 2013
12
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
Jan 17, 2013
1,667
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

Administrator
Staff member
Feb 24, 2012
13,775
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Hello! You can do it using the quotient rule too. Perhaps you missed something in the calculations. :) Here is how they go:

$$\frac{d}{dr} \frac{a}{1-r} = \frac{ \left( \frac{d}{dr} (a) \right) (1-r) - (a) \left( \frac{d}{dr} (1-r) \right)}{(1-r)^2} = \frac{0 - a(-1)}{(1-r)^2} = \frac{a}{(1-r)^2}.$$

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. ;)

Cheers! :D