Results 1 to 11 of 11

Math Help - Another derivative question

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    10

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Your answer to the first question is correct, but ONLY if you put in parentheses like so:

    (x^2+4x+24)/(x+2)^2. To use a better-looking format, your answer is this:

    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2010
    Posts
    10
    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) ?
    Is this the right answer?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Closer. However, you have not taken the derivative of the argument of the logarithm function correctly. What's going on inside the logarithm?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    10
    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)?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2010
    Posts
    10
    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,
    the answer would be..
    1/(x tan x) (tan x + x sec^2 x)
    Is this right?
    Btw, thank you verrry much for the help again
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Oct 2010
    Posts
    10
    Thanks alot Ackbeet.
    I really get it now
    You are my master!
    Thank you!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You're welcome for the help, but please don't make me your master.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative question.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 14th 2011, 11:15 PM
  2. derivative question.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 15th 2010, 11:06 AM
  3. Replies: 19
    Last Post: October 19th 2009, 06:10 PM
  4. Derivative question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 24th 2007, 03:33 PM
  5. Derivative Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2006, 04:25 PM

Search Tags


/mathhelpforum @mathhelpforum