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

I know that's not right...please help.

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- Aug 7th 2006, 07:37 PMbeckyderivatives
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

I know that's not right...please help. - Aug 7th 2006, 07:47 PMThePerfectHackerQuote:

Originally Posted by**becky**

Now,

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

Thus, using the generalized power rule,

$\displaystyle (x+1)^{-2}$ - Aug 7th 2006, 07:47 PMSoroban
Hello, becky!

Quote:

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

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

What's wrong with using the Quotient Rule?

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

- Aug 7th 2006, 07:50 PMSoltras
Hi Becky

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

So while $\displaystyle \frac{x+1}{x}=\frac{x}{x}+\frac{1}{x}$, it's not true that $\displaystyle \frac{x}{x+1} = \frac{x}{x}+\frac{x}{1}$.

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 $\displaystyle g$ and $\displaystyle h$ are differentiable functions, then $\displaystyle \left(\frac{g}{h}\right)' = \frac{g'h-h'g}{h^2}$.

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

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

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

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

$\displaystyle =\frac{1}{(x+1)^2}$ - Aug 7th 2006, 08:06 PMbeckyThanks for your help
You all explain it so much better than a textbook.