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Trying to use product rule on complicated function

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    This is the problem and my attempt. I understand that I can take ln of both sides. However, I am trying to see if it is at all possible to arrive at the correct derivative without taking ln of both sides.

    I am under the impression that an alternative to taking ln of both sides is to use the power rule on terms with a variable as the exponent. So using this approach to this function, what am I doing wrong? If all is correct so far, I can not see what step to take next.


    productrule1-1.png
     
    Last edited: Mar 4, 2012
  2. jcsd
  3. Mar 4, 2012 #2

    Mark44

    Staff: Mentor

    It's easier if you don't use radicals. Instead use fractional exponents, and then change back from fractional exponents to radicals at the very end, if needed.
     
  4. Mar 4, 2012 #3
    Hi Mark, thanks for responding to my question. I tried evaluating this function by using fractional exponents instead and I end up in the same spot. Where am I going wrong here? Or if all is correct so far, what would be the next logical step? I can't think of anything here.

    EDIT: let me double check again
     
  5. Mar 4, 2012 #4
    Try labeling all of your functions as f(x), g(x), etc, figuring out those derivatives, and then solving it in terms of those functions. that is, let f(x) = x^2-1, g(x) = x^2 + 1, and g(x) = e^root(x). figure out all those derivatives in a little table form. it's easier to keep organized.
     
  6. Mar 4, 2012 #5
    Mark, I had to double check my work using the method you suggested. I seem to be winding up at the same spot and getting stuck there. Am I doing something wrong here? If not, what should I do from here?


    productruleb.jpg
     
  7. Mar 4, 2012 #6
    The derivative of e^x^1/2 is not ln e* e^x^1/2. Use the chain rule here.

    (edited)
     
    Last edited: Mar 4, 2012
  8. Mar 4, 2012 #7
    80past, I didn't write (lne^x^1/2) though, I wrote (lne)(e^x^1/2), which is (1)(e^x^1/2), which is e^x^1/2.

    I was told that you can't use the chain rule or power rule on terms with a variable as a radical..and that you must use the exponential rule, is this not correct?


    Edit: Even in light of the correction, I was under the impression that when you have a variable as an exponent, you must either take ln of both sides or use the exponential rule, which is why I decided to use the exponential rule. Is there an exception for the number e?
     
  9. Mar 4, 2012 #8
    Yeah, I made an error and edited it to make it match what you said. But it is still incorrect. I'm a little bit confused as to being told not to use the chain rule with a variable as a radical... The derivative of e^u is e^u*u'.

    you wouldn't use the power rule, because that's when the variable is the base, not the exponent, but yes, you are using the exponential rule. But because you have a function of x, you need to take that into account.

    For example, what's the derivative of e^2x?
     
  10. Mar 4, 2012 #9
    No, there's no exceptions that I'm aware of. Look up the chain rule.
     
  11. Mar 5, 2012 #10

    Mark44

    Staff: Mentor

    d/dx(eu) = eu * du/dx
    You can apply this when you differentiate ex1/2

    Your other derivative is incorrect as well. You have (1/3)[(x2 - 1)/(x2 + 1)] times some other stuff, but you forgot to decrease the exponent by 1.
     
  12. Mar 5, 2012 #11

    Recent discussion with Office_Shredder


    I had a constant to the power of a variable:

    s(t) = .5^t

    The power rule is: s' = t(.5^(t-1))

    however, that can't be done because it is a shortcut of the chain rule, and the full version of the chain rule yields:

    s' = t(.5^(t-1))*(.5)' = 0

    however, s' ≠ 0

    so chain rule can't be done. Is this correct?

    Then it would be that the derivative of f(x) = e^x

    f' = x(e^(x-1)*e' = 0, which is not correct. So the chain rule can't be used.


    However, you ask what the derivative of e^2x is, and the chain rule seems to work fine on that example, yielding 2e^2x.

    So the chain rule works on fx = e^2x, and works for fx = e^x

    Thus, its becoming apparent that there are two different ways of applying the chain rule. One way is bringing the exponent down to the coefficient and subtracts the exponent by 1 and then multiplies by the derivative of the coefficient (in otherwords, the derivative of the outside times the the derivative of the inside). Such as would be the case with fx =x^2. And the other way of applying the chain rule is to use substitution of the variable such as would be the case for taking e^x.

    So you are saying the chain rule can be used, it just depends on how you apply the chain rule?
     
  13. Mar 5, 2012 #12
    Ah okay, i'm going to make those corrections and see how it goes
     
  14. Mar 5, 2012 #13
    you are looking at several different rules at once. in one point, you mention mixing up the power rule and the exponential rule. that is because f(x)=x^2 is a different type of function than f(x) = 2^x. in the first, you have a variable being raised to a constant power, and in the second, you have a constant being raised to a variable.

    the chain rule says f(g(x))' = f'(g(x))*g'(x). this is true for any function.

    the first method you describe is the power rule, albeit a little muddied. the power rule in words (so much easier in symbols): you have a variable raised to a constant power; take that constant power and multiply the variable, raised to the power of the original constant minus 1. (in symbols: if f(x) = a*x^n, then f'(x) = a*n*x^(n-1). you don't necessarily need to use the chain rule for the second case (although you actually are); you are using the exponential rule, but the derivative of x is 1, so in a way you are using the chain rule. specifically, if f(x) = e^x, then f'(x) = ln e * e^x (d/dx) = ln e*e^x*1 = e^x.

    what do you mean a shortcut of the chain rule? you can't do it because you should be using the exponential rule, not the power rule.

    i guess the chain rule must be used where it's needed (a bit of a tautology). for a long time, i didn't like using the quotient rule, so i would always turn a quotient into a product/chain rule problem, which seemed a little easier to me. but look at your definitions and your rules. remember exactly where to use each one. you can follow the rules pretty blindly (assuming you're accurately following the rules) and still get the right answers. that's why earlier i mentioned writing everything out as functions (e.g. f(x) instead of x^2-1 and making a table of your functions and their derivatives). sometimes it's easier to apply these rules without the "real" functions being there muddying up the problem.
     
    Last edited: Mar 5, 2012
  15. Mar 5, 2012 #14
    Okay, this is what I have so far...starting to get real close. I just can't figure out the last part of 2x/((x^2)-1)-((x^2)+1)

    productrule3.jpg


    80past2, thanks for clearing some things up. I'm getting a better handle on how the various derivative rules apply to various types of functions. And i'm going to have to look further into rewriting functions with tables as you suggest as i'm not totally sure of how i'd go about evaluating in such a way.
     
  16. Mar 5, 2012 #15
    well, imagine that you didn't know what the functions were. they can be anything. so now i ask you, what is the derivative of (f(x)/g(x))^(1/3)*h(x).
     
  17. Mar 5, 2012 #16
    Here's what I got when I took the derivative of (f(x)/g(x))^(1/3)*h(x).

    Can you show me where, in post #14, i'm going wrong and why I'm getting (2/3)(2x/(x[itex]^{2}-1[/itex])(x[itex]^{2}+1[/itex]) instead of (2/3)((x/(x[itex]^{2}-1[/itex])-(x/(x[itex]^{2}+1[/itex])))


    newmethod.jpg
     
  18. Mar 5, 2012 #17
    hint: simplify (into one fraction) (x/(x^2−1)-(x/(x^2+1)))
     
  19. Mar 5, 2012 #18

    fractions.jpg

    Okay, the answer I arrived at using product rule is equivelent to the final answer in post #1, which was arrived at by using the exponential rule, just in a different form, is the correct? If so, then is it possible to go the reverse direction, from 2x/((x^2)-1)-((x^2)+1) to (x/(x^2−1)-(x/(x^2+1)))?
     
  20. Mar 5, 2012 #19
    Well, if the two are equal, then you can certainly go between them. Just go backwards from what you did to get the result they want. I'm still confused about what you mean by using the product rule and exponential rule. You don't have much of a choice of when you get to use one or the other. You use the power rule if you have a power function, and you use the exponential rule if you have an exponential function.
     
  21. Mar 5, 2012 #20

    Okay, so a power function means a function like y = x^4, and an exponential function means a function like y = 4^x.

    If I have a power function, I can use the power rule, chain rule, or even take log of both sides, and arrive at the correct answer. However, I can not use the exponential rule.

    If I have an exponential function, I can not use the power rule, can not use chain rule. However I can take ln of both sides, then derive. And I can use the exponential rule.

    If I have the special case exponential function, e^x, then I can then use chain rule (by u substitution of x), or take ln of both sides, or use the exponential rule. However, I can not use the power rule.

    I think i've been mixing these rules up a bit over this and most recent thread previous to this one. But this is my understanding as of now. I think its correct, but not entirely sure.


    EDIT: Note, just before making this post, I created a new thread specifically addressing the confusion with the various derivative rules: https://www.physicsforums.com/showthread.php?t=584204
     
    Last edited: Mar 5, 2012
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