Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - Question on derivative question involving quotient & chain rule

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    229

    Question on derivative question involving quotient & chain rule

    There are 2 questions involving derivatives that I have trouble getting to the answer for. The first one is y = [(6-5x) / (x^2 - 1)]^2

    and the 2nd one is

    y = 36 / (3-x)^2

    The answers respectively are;
    [2(6 - 5x)(5x^2 - 12x + 5)] / (x^2 - 1)^3
    &
    y = (8/3)x + 4

    It would be very much appreciated if anyone could provide step-by-step solutions to both problems.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Just in case a picture helps...



    ... where



    ... is the chain rule, and...



    ... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).

    So what you've got here is two chains inside a product-rule (the legs-uncrossed version). That's just my preference - you could do it as a product (or even a quotient - yeughh!) inside a chain.

    Anyway, if you feel up to filling in the blanks and simplifying, notice I've tweaked the third balloon along for the sake of the common denominator.

    Hope that helps, or doesn't further confuse.




    __________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus: Gallery

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; October 19th 2009 at 03:40 PM. Reason: third not second
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    229
    Do you have a solution for the 2nd problem by any chance?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Hate to say this (after last time) but are you QUITE sure that's the right derivative?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    229
    For the 2nd question?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Oh yes
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    229
    Oh damn, the answer must have confused you because it's not the answer to the derivative directly, but an equation where the point is (0,4).

    Because the question is to find an equation of the tangent line to the graph of the function at the given point. I'm terribly sorry for overlooking that!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Oh, fine. The derivative is the first step, it's not a hard one, just the chain rule in a simple way. I'll post a pic just in case it helps, but go ahead and try it, then I guess you know to plug in zero as the x-value in the derivative function, to find the slope of the tangent line...



    No need to simplify or anything, just plug zero (as the value of x) into the derivative and you have the slope of the line. Now, the y-value where x is zero has a special role in one form of the line's equation, no?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    Posts
    229
    Oh yeah, the second part is easy. It's only the derivative part that I'm concerned about.

    Could you please include all the steps for the first question? I don't know how they got to that long answer, mine doesn't match us.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Well, can you see that the denominators match?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Put it another way... can you see why the final (correct) denominator is (x^2 - 1)^3
    ?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2009
    Posts
    229
    My denominator matches that of the answer, but I must've messed up somewhere on the calculations of the numerator because they don't match.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    OK, you're ahead of me, so I don't know for sure if the numerator will match*, but do you get (for the numerator) the following...

    2(6 - 5x)(-5)(x^2 - 1) + (6 - 5x)^2 (-2) (2x)

    If not, read off from the bottom row of balloons


    *Yep, it does
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Oct 2009
    Posts
    229
    Quote Originally Posted by tom@ballooncalculus View Post
    OK, you're ahead of me, so I don't know for sure if the numerator will match, but do get (for the numerator) the following...

    2(6 - 5x)(-5)(x^2 - 1) + (6 - 5x)^2 (-2) (2x)

    If not, read off from the bottom row of balloons
    I got all of the above values except the (x^2 - 1) + (6 - 5x)^2. How did you get these two?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Well, I can only suggest checking yours against the bottom row of balloons and checking that the balloons are following the product-rule, as they should. The product rule may be hard to see amongst all the chain rule details so I might simplify...

    By the way, beware the 'tweak', third balloon along, which is for the sake of the common denominator.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Derivative Help using Quotient & Chain Rule
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 5th 2011, 04:23 PM
  2. Chain Rule Derivative Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 13th 2011, 01:59 PM
  3. simple question - derivative - quotient rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 23rd 2010, 05:10 PM
  4. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  5. Derivative question using the Quotient rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 17th 2007, 09:05 AM

Search Tags


/mathhelpforum @mathhelpforum