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Homework Help: How do you differentiate a fraction with x in the numerator and denominator? :/

  1. Jan 8, 2013 #1
    1. Find derivative:

    g(x) = (e^x)/(2 + 3x)

    2. The fraction confuses me...
    I wrote it instead as: (e^x)(3x^-1)(2^-1)

    2^(-1) equals .5, so then I thought it was: (e^x)(1.5x^-1)

    And then I thought it's time to differentiate...

    So I got: (-1.5x^-2)(e^x)

    And then I thought, " Since these are multiplied together, do you just differentiate the e^x and it still ends up as (1.5x^-1)(e^x)...?

    But none of these ideas are right... =_=
    Please help me!
    Thank you for your help!
  2. jcsd
  3. Jan 8, 2013 #2


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    You don't have calculus problems above all, you have basic algebra problems. (2+3x)^(-1) means 1/(2+3x). That's not the same as 1/2+1/(3x) or 2^(-1)+(3x)^(-1). Just try using the quotient rule on the given expression.
  4. Jan 9, 2013 #3
    Unfortunately, basic algebra gives me troubles... =_=

    So when I use the product rule, it turns out to be:

    (2 + 3x)(e^x) - (e^x)(3x^-1)/ (2 + 3x)^2

    And that is:

    2e^x + 3xe^x - 3x^-1e^x / 9x^2 + 12x + 4

    Can this be simplified any more?

    If you tried to move the -3x^-1*e^x to the denominator, would you get:
    -3x*e^x or would you get -3x*e^-x?

    Please help me with my basic algebra! D:
  5. Jan 9, 2013 #4
    The reason I ask that is because if it was:

    2e^x + 3xe^x / 9x^2 + 12x + 4 - 3xe^x, then wouldn't the 3xe^x in the numerator and denominator cancel, leaving us with a -1?

    And then it's: 2e^x -1 / 9x^2 +12x + 4

    Basic algebra! HELP!
  6. Jan 9, 2013 #5


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    If you want to use Product Rule, then you can simply try

    Since you already have a numerator and denominator arrangement as the function, you can choose to use the Quotient Rule.

  7. Jan 9, 2013 #6


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    You hit a calculus problem before your algebra problems. The derivative of (2+3x) is NOT 3x^(-1). What rule are you trying to apply here?
  8. Jan 9, 2013 #7


    Staff: Mentor

    No, that's not right, either. You're missing the -1 exponent on the first (2 + 3x) factor.

    We can show you where you're going wrong, but you're going to need to get yourself up to speed on your own. If you're having problems with basic algebra, and you're trying to do calculus, that's a recipe for disaster.
  9. Jan 9, 2013 #8


    Staff: Mentor

    No. That's not how it works.

    Do you think that $$ \frac{5 + 1}{2 + 5} = 1/2?$$

    The only time you can cancel is when you have the same factors (expressions being multiplied) appearing in the numerator and denominator. You CAN'T cancel if you have the same terms (expressions being added) in numerator and denominator.
  10. Jan 9, 2013 #9
    Okay. So now I'm just going to try to use the Product Rule.

    g(x) = e^x/(2 + 3x)

    e^x(2 + 3x)^-1

    (e^x * -1(2 + 3x)^-2 * 3) + (e^x)(2 + 3x)^-1

    e^x * -3(2 + 3x)^-2) + (e^x/2 + 3x)

    (-3e^x /(2 + 3x)^2) + (e^x/2 + 3x)

    (-3e^x / 9x^2 + 6x + 4) + (e^x)^2/(2 + 3x)^2

    (-3e^x/ 9x^2 + 6x + 4) + (e^(2x) / 9x^2 + 6x + 4

    = -3e^x + e^(2x) /(9x^2 + 6x +4)

    ......So this is what I get when I try to follow the product rule...
    Is this what the answer is supposed to be?
    Am I still doing something wrong? Help.
    Thank you for the help.
  11. Jan 9, 2013 #10
    For some reason in my earlier posts I said "product rule" when I really meant "quotient rule..."

    My method of quotient rule is not right...?
  12. Jan 9, 2013 #11
    @Dick - Oh, wow. Thanks for pointing out that crucial mistake! :/

    When I was trying to do the quotient rule, I said the derivative of of 2 + 3x was 3x^-1!

    No, the derivative of that is 3!

    So, now it's:

    (2 + 3x)(e^x) - (e^x)(3)/(9x^2 + 6x + 4)

    2e^x + 3xe^x - 3e^x / (9x^2 + 6x + 4)

    = -e^x + 3xe^x / 9x^2 + 6x + 4

    ...So, that's a different answer from when I did the product rule! Why?!
    I am doing something wrong...
    Please help.
    Thank you!
  13. Jan 9, 2013 #12
    OKAY! UGH!

    So the computer program I am working on counted the answer: -e^x + 3xe^x/ 9x^2 + 6x + 4 as wrong.

    The right answer had to be in the format: -e^x + 3xe^x / (2 +3x)^2
    I got the points for the problem, though, so that's good.

    BUT I still want to know where I went wrong when I was trying to do the product rule!
  14. Jan 9, 2013 #13


    Staff: Mentor

    I can't believe that the program thought the above was correct. What you showed here is considered to be
    $$ -e^x + \frac{3xe^x}{(2 + 3x)^2}$$

    The correct answer, in inline text, needs parentheses around the terms in the numerator, like so:
    (-e^x + 3xe^x) / (2 +3x)^2
  15. Jan 9, 2013 #14


    Staff: Mentor

    As already mentioned, you need more parentheses.

    This line -- 2e^x + 3xe^x - 3e^x / (9x^2 + 6x + 4) --
    should be written as (2e^x + 3xe^x - 3e^x) / (9x^2 + 6x + 4)

    This line -- -e^x + 3xe^x / 9x^2 + 6x + 4 --
    should be written as (-e^x + 3xe^x) / (9x^2 + 6x + 4)

    Also, I don't see any advantage in expanding (3x + 2)2 as you did.

    Using the product rule you end up with the sum of two rational expressions. Using the quotient rule, you end up with a single rational expression. In your work with the product rule, get a common denominator and combine the two expressions, and it should simplify to the same as what you got with the quotient rule.
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