quotient rule differentiation troubles

[lmerry]

okay, i don't usually have any trouble with this but i just cant seem to make my work match what my calculator says the answer is. So here we go.

Find the first derivative:

then, i try to simplify:

(there is a typo in the above image, it's supposed to be a negative x on the top, and a positive 1 on the bottom of the last equation, obviously)

the answer my TI-89 calculator spits out when i use the differentiate function is

f'(x)=-2(x-1)

and i can't get it. please tell me where i screwed up, or if im just out of it and not thinking of some simple simplification type of thing to get from where i am to the "correct" answer.

i'd also like to know what some of you use to type math problems for this forum, ive been using a program called MathType for Mac, but the trial is about up. Its also quite a pain to type it, capture it as an image with grab, convert from tiff to jpg, upload to photobucket, and then add the link. is there any easier way?

 

[stapel]

Going from your next-to-last to your last line in your computations, how did you get that (-2)/(2), in the numerator, equalled 0? (The constant term disappears, is why I ask.)

Thank you.

Eliz.

 

[soroban]

Hello, lmerry!

Find the first derivative: \(\displaystyle \L\,f(x) \:=\:\frac{-x^3\,+\,x^2\,+\,3x\,+\,1}{x\,+\,1}\)

I got: \(\displaystyle \L\:f'(x) \:=\:\frac{-2(x^3\,+\,x^2\,-\,x\,-\,1)}{(x\,+\,1)^2}\)

My TI-89 says: \(\displaystyle \:f'(x)\:=\:-2(x\,-\,1)\)

Did you try to reduce your answer?


The numerator has: \(\displaystyle \:\underbrace{x^3\,+\,x^2}\,-\,\underbrace{x\,-\,1}\)

Factor in pairs: . . . \(\displaystyle x^2\underbrace{(x\,+\,1)}\,-\,\underbrace{(x\,+\,1)}\)
. . . . . . . . . . . . . . . . . .\(\displaystyle \downarrow\;\;\swarrow\)
Common factor: . . . .\(\displaystyle (x\,+\,1)\,(x^2\,-\,1)\)

Diff. of squares: . .\(\displaystyle (x\,+\,1)\overbrace{(x\,+\,1)(x\,-\,1)}\)


Therefore: \(\displaystyle \L\:f'(x)\;=\;\frac{-2\sout{(x\,+\,1)^2}(x\,-\,1)}{\sout{(x\,+\,1)^2}} \;=\;-2(x\,-\,1)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can cancel \(\displaystyle (x\,+\,1)\) because \(\displaystyle x\,\neq\,-1.\)

. . In the original function: \(\displaystyle \,f(-1)\:=\:\frac{0}{0}\)


In fact, this tells us that \(\displaystyle f(x)\) can be reduced . . . Ha!

. . \(\displaystyle \L f(x) \:=\:\frac{-x^3\,+\,x^2\,+\,3x\,+\,1}{x\,+\,1}\:=\:\frac{(x\,+\,1)(-x^2\,+\,2x\,+\,1)}{x\,+\,1}\)


Hence: \(\displaystyle \L\:f(x) \:=\:-x^2\,+\,2x\,+\,1\) . . . for \(\displaystyle x\,\neq\,-1\)

No wonder that: \(\displaystyle \L f'(x)\:=\:-2x\,+\,2\) . . . Very funny (not!)