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Product Rule Problems

  1. Feb 3, 2016 #1
    • Member warned about posting with no effort shown
    1. The problem statement, all variables and given/known data
    7ba031cc28ef8d4157ac40ec1b6faf93.png

    2. Relevant equations
    The product rule formula.

    3. The attempt at a solution
    I managed to solve 45/50 product rule but I can't seem to solve these ones. Apparently you use product rule to solve these.
     
  2. jcsd
  3. Feb 3, 2016 #2

    SteamKing

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    These look like integration problems. The "product rule" refers to finding the derivative of the product of two or more functions.

    You can use the product rule on each answer to find f(x), but to go the other way, either u-substitution or integration by parts will probably be required.

    Still, you'll have to show your attempts at solving these problems in order to get help.
     
  4. Feb 3, 2016 #3

    SammyS

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    From those answers, it's clear that you're finding the anti-derivative of f(x).

    Consider the first problem. Rewrite the given function.
    ##\displaystyle f(x)=\frac{x-1-x\ln x }{x(x-1)^2 } \ ##
    ##\displaystyle =\frac{x-1 }{x(x-1)^2 } - \frac{x\ln x }{x(x-1)^2 } \ ##
    ##\displaystyle =\frac{1}{x}(x-1)^{-1} - (x-1)^{-2}\ln x \ ##​

    Apply the product rule in reverse.
     
  5. Feb 3, 2016 #4
    At the University I go to Integration by parts is also known as the product rule, even if that terminology is not correct this is what I am use to for naming convention. Thank you for the help with that problem. The problem I'd like to tackle next is the second problem.

    Here is my solution:
    ∫ ln(lnx) + 1/(lnx)^2
    Obtain a common denominator:
    ∫ (ln(lnx)(lnx)^2)/(lnx)^2+ 1/(lnx)^2

    Unsatisfactory solution:

    Attempt a u substitution

    let u = lnx
    du = 1/x dx

    no 1/x present therefore not satisfactory.

    Looking for suggestions.

    Thank You!

    PS. How does one use the built in equation editor on this website. I've been looking for a while
     
  6. Feb 3, 2016 #5

    SteamKing

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    You can get some symbols and Greek letters by hitting the big ∑ sign in the blue toolbar across the top of the edit box. These are used for simple expressions.

    For more complex equation editing, PF uses Latex, a guide to which can be found by hitting the LaTeX/BBcode Guides button at the lower left hand corner of the edit box.
     
  7. Feb 3, 2016 #6

    SteamKing

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    Your u-substitution did not go correctly because you overlooked the fact that you also have a ln(ln x) term in your integral.

    If you let u = ln x, what does ln(ln x) become?
     
  8. Feb 3, 2016 #7
    this becomes ln(u) however how does dx become du?
     
  9. Feb 3, 2016 #8

    SteamKing

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    If u = ln (x), what would you do to this equation to obtain x by itself?
     
  10. Feb 4, 2016 #9
    x = 10^u but I do not see the benefit to such a manipulation?
     
  11. Feb 4, 2016 #10

    Mark44

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    No, that's not correct. The inverse of the natural log function (ln(x)) is the natural exponential function, ex, not 10x.

    Integration by parts is using the product rule in reverse, but the term "product rule" is pretty much universally used to describe a differentiation rule. I.e., ##\frac {d}{dx} (fg(x)) = f(x) * g'(x) + f'(x) * g(x)##.
     
  12. Feb 4, 2016 #11
    I recognize the difference between the two I'm just making note of how my teacher has described the rule in class. If you analyze the term "product rule" using the socratic method one may deduce that a term can never truly be perfect. Perhaps this could be called a product rule of integrals, since it does involve product rules.

    Edit: I suppose arguing about a definition is beyond the point of this post. Furthermore I will transfer some of my attempts to the solution section.

    Also even if x = e^u I still do not see the benefit to such a approach. Is it possible to get some more information?

    Sincerely,

    OmniNewton
     
  13. Feb 4, 2016 #12

    SteamKing

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    Not until you've gone further in your solution attempts than you have shown so far.

    You haven't even written the complete, original integral out using the substitution u = ln(x).
     
  14. Feb 4, 2016 #13
    I'm sorry I don't understand how you guys use substitutions on these forums. Strangely whenever I ask about these it seems that you guys will try to integrate without changing dx to du. How can I integrate if not all variables are u and we are not integrating with respect to u. For example, if you have ∫x^2cos(x^5) dx one cannot simply say let u = x^2 so ∫ucos(u^3) dx One cannot integrate unless dx is changed to du correct?
     
  15. Feb 4, 2016 #14

    SteamKing

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    If you make a substitution of u = x2, then du = 2x dx by the chain rule.

    A substitution of u = x2 may not be the best one for the particular integral in your example.

    This link gives a good explanation of the mechanics of u-substitution:

    http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinite.aspx
     
  16. Feb 4, 2016 #15

    Mark44

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    The term "product rule" in the context of a course in calculus is widely understood to refer to the product rule in differentiation.

    Regarding "using the socratic method one may deduce that a term can never truly be perfect." -- language often doesn't work this way. Language expressions can have meanings that are commonly understood, that are different from literal meanings of the words in the expression. Parsing a simple expression such as "good bye" using the Socratic method doesn't get you very far.
     
  17. Feb 4, 2016 #16

    Mark44

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    Correct. If ##u = x^2## then ##du = 2xdx##, or ##dx = (1/2)\frac {du}{x}##. But ##\int x^2 \cos(x^5)dx \ne \int u \cos(u^3)du##. Using this substitution, ##x^5## does not become ##u^3##. And there's also the matter of dx changing to du.
     
  18. Feb 4, 2016 #17
    Sorry maybe I'm misunderstanding however, I was told u-substitution only works if a factor is the derivative of what you are substituting. I do not see this in the context of my original problem
     
  19. Feb 4, 2016 #18

    SteamKing

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    Some substitutions are easy to see, others are not so obvious. You often don't know if a particular substitution works until you grind out the algebra and hit a wall.
     
  20. Feb 4, 2016 #19
    let u = lnx
    du = 1/x dx
    ∫ ln(u) + 1/(u)^2 dx

    see from this point I do not understand where to go? As I still have the dx
     
  21. Feb 4, 2016 #20

    SteamKing

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    But you are also ignoring the transformation eu = x.
     
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