# Thread: Very basic derivatives: 1/(1-x) and x/(1-x)

1. ## Very basic derivatives: 1/(1-x) and x/(1-x)

I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
$
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}
$

$
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}
$

But two derivatives can be the same only if the functions differ only with a constant!

I know, I must be making some terribly stupid mistake somewhere. But I can't find it.

2. Originally Posted by lampak
I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
$
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}
$

$
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}
$

But two derivatives can be the same only if the functions differ only with a constant!

I know, I must be making some terribly stupid mistake somewhere. But I can't find it.
$\frac{1}{1-x}-1 = \frac{1}{1-x}-\frac{1-x}{1-x}=\frac{x}{1-x}$

3. Originally Posted by lampak
I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
$
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}
$

$
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}
$

But two derivatives can be the same only if the functions differ only with a constant!

I know, I must be making some terribly stupid mistake somewhere. But I can't find it.
$\frac{1}{1-x}$ and $\frac{x}{1-x}$ differ by a constant.

$\frac{x}{1-x} = \frac{x+1-1}{1-x} = \frac{1 - (1-x)}{1-x} = \frac{1}{1-x} - 1$

4. Great thanks for quick reply