# derivatives

• Aug 7th 2006, 08:37 PM
becky
derivatives
I always have difficulties with fractions...

Find the derivative of f(x) = (x/x+1) - 3

I usually break it down to f(x) = x/x +x/1 -3

f'(x) = 1 + 1

• Aug 7th 2006, 08:47 PM
ThePerfectHacker
Quote:

Originally Posted by becky
I always have difficulties with fractions...

Find the derivative of f(x) = (x/x+1) - 3
.

You can forget about -3 because it is a constant.

Now,
$\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}=1-(x+1)^{-1}$
Thus, using the generalized power rule,
$(x+1)^{-2}$
• Aug 7th 2006, 08:47 PM
Soroban
Hello, becky!

Quote:

Find the derivative of: $f(x) \:= \:\frac{x}{x+1} - 3$

I usually break it down to: $f(x) \:= \:\frac{x}{x} + \frac{x}{1} -3$ . . . I hope not!

What's wrong with using the Quotient Rule?

$f'(x)\;=\;\frac{(x+1)\cdot1 - x\cdot 1}{(x+1)^2} - 0 \;= \;\frac{1}{(x+1)^2}$

• Aug 7th 2006, 08:50 PM
Soltras
Hi Becky

You can only break up a fraction if the added terms are on top, not bottom.

So while $\frac{x+1}{x}=\frac{x}{x}+\frac{1}{x}$, it's not true that $\frac{x}{x+1} = \frac{x}{x}+\frac{x}{1}$.
So we can't use that kind of techniue in this particular situation.

What you want to do here is apply the quotient rule, which basically says that if $g$ and $h$ are differentiable functions, then $\left(\frac{g}{h}\right)' = \frac{g'h-h'g}{h^2}$.

In your case, $g$ is the top of the fraction $x$ and $h$ is the bottom, $x+1$.

So $g'=1$ and $h'=1$.

Then $f'(x) = \frac{g'h-h'g}{h^2}-0$

$=\frac{1\cdot(x+1)-1\cdot(x)}{(x+1)^2}$

$=\frac{1}{(x+1)^2}$
• Aug 7th 2006, 09:06 PM
becky