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.
Hello, becky!
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?
$\displaystyle f'(x)\;=\;\frac{(x+1)\cdot1 - x\cdot 1}{(x+1)^2} - 0 \;= \;\frac{1}{(x+1)^2}$
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 $\displaystyle f'(x) = \frac{g'h-h'g}{h^2}-0$
$\displaystyle =\frac{1\cdot(x+1)-1\cdot(x)}{(x+1)^2}$
$\displaystyle =\frac{1}{(x+1)^2}$