# Another derivative question

• Oct 18th 2010, 07:37 AM
kunkka
Another derivative question
Sorry, I just couldn't find out by my self :(

Anyone got a clue for these three question?

f(x) = (x^2 - 24) / (x+2)

y= ln(x tan x)

I know it looks very simple but I'm totally lost.

Please give me a hint for those two ,

Thank you.(Smirk)
• Oct 18th 2010, 08:03 AM
Ackbeet
For these sorts of problems, you have to kind of "reverse the order" of evaluation. Take f(x) = (x^2 - 24) / (x+2). The last arithmetic operation you would do in evaluating the function at a point would be the division. Therefore, that's the first thing you differentiate. What is the quotient rule?

For the second function, you work from outside in. You'll need the chain rule as well as the product rule. What do you get?
• Oct 18th 2010, 08:11 AM
kunkka
For the 1st question, I used quotient rule and got
x^2+4x+24/(x+2)^2
Hope it's right, thanks. :)

For the 2nd question,
Would it be same as (ln x)(ln tan x)???
I have no idea how to start this.
• Oct 18th 2010, 08:23 AM
Ackbeet
Your answer to the first question is correct, but ONLY if you put in parentheses like so:

$x^2+4x+\dfrac{24}{(x+2)^{2}},$ whereas the correct answer is this:

$\dfrac{x^{2}+4x+24}{(x+2)^{2}}.$ They are not the same!

For the second problem, I don't think you understand yet. What is the chain rule? How does it apply here? What is the outermost function?
• Oct 18th 2010, 08:30 AM
kunkka
Ok, thanks for the help.
To start off, would it be..
let's say u = ln(v) and v = (x tan x)
1/(x tan x) (1 sec^2 x) ?
• Oct 18th 2010, 08:33 AM
Ackbeet
Closer. However, you have not taken the derivative of the argument of the logarithm function correctly. What's going on inside the logarithm?
• Oct 18th 2010, 08:35 AM
kunkka
In the logarithm, I can only think of the first x becoming to be 1 after the derivative, and tan x becoming sec^2 x.. :(
Would there be any other method to solve (x tan x)?
• Oct 18th 2010, 08:42 AM
kunkka
Oh, I used product rule on (x tan x).
As a result I got
(tan x + x sec^2 x)
Therefore, to substitute this into the chain rule,
1/(x tan x) (tan x + x sec^2 x)
Is this right? :)
Btw, thank you verrry much for the help again :)
• Oct 18th 2010, 09:04 AM
Ackbeet
Yeah, that's looking better. I would probably write the answer as

$\dfrac{\tan(x) + x \sec^{2}(x)}{x\tan(x)}.$

Then it's perfectly clear what is in the argument of each trig function, and what's in the numerator and denominator of the fraction.
• Oct 18th 2010, 09:07 AM
kunkka
Thanks alot Ackbeet.
I really get it now :)
You are my master!
Thank you!
• Oct 18th 2010, 09:07 AM
Ackbeet
You're welcome for the help, but please don't make me your master.