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Computing derivative

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    y = x 2sinx

    2. Relevant equations


    3. The attempt at a solution
    Ok, so If I see an x in an exponent, I would want to use ln to 'bring it out', right?

    ln y = ln (x 2sinx) = ln x + ln 2sinx = ln x + sinx ln2

    now I take the derivative :

    y'/y = 1/x + cosx ln2

    multiply both sides by y

    y' = [1/x + cosx ln2] (x 2sinx)

    is everything correct?
     
  2. jcsd
  3. Mar 1, 2015 #2
    Yes it looks correct. You could have also done a product rule + chain rule expansion. That way you could avoid the implicit form,
     
  4. Mar 1, 2015 #3
    ok, cool. Thanks!
     
  5. Mar 1, 2015 #4

    SteamKing

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    Is x multiplied by 2sin x, or is it something else?
     
  6. Mar 1, 2015 #5
    yes , multiplied.
     
  7. Mar 1, 2015 #6
    could you show your work for that?
     
  8. Mar 1, 2015 #7

    Ray Vickson

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    It is correct, but should be re-written as ##y' = 2^{\sin(x)} + \ln(2)\, \cos(x)\, x \, 2^{\sin(x)}##. The way YOU wrote it would not allow you to easily get ##y'(0)## because it would involve the "illegal" fraction ##\frac{0}{0}##. The way I wrote it presents no problems at ##x = 0##.
     
  9. Mar 1, 2015 #8
    y = x 2sinx

    Product rule says [itex]y'(x) = f'(x)g(x) + f(x)g'(x)[/itex]. Choose f(x) = x, and g(x) = 2sinx. (although order doesn't really matter)

    Now, all you have to do is compute the derivative of f, the derivative of g, and then plug. The derivative of g also involves the chain rule. Try figuring it out for yourself, it should be more rewarding. You already know the answer so you'll know if you've got it. Good luck :D
     
  10. Mar 1, 2015 #9
    Wait, how did you get 2sin(x)+ln(2)cos(x) when I got 1/x + ln2 cosx?
     
  11. Mar 1, 2015 #10
    but to get the derivative of 2sinx, I need to use ln right? is there any other way?
     
  12. Mar 1, 2015 #11
    He just distributed the term you had outside your parenthesis.

    The derivative of some scalar a, raised to the x, is:
    [itex] \frac{d}{dx}[a^x] = (a^x)ln(a)[/itex] ... as a side note unrelated to this problem, think about when the scalar a = e.
    So, applying the chain rule the derivative of scalar a, raised to some f(x), is:
    [itex] \frac{d}{dx}[a^{f(x)}] = (a^{f(x)})ln(a)f'(x)[/itex]
     
  13. Mar 1, 2015 #12
    scalar a = constant right?

    also, this whole time, i thought [itex] \frac{d}{dx}[a^x] = (x)ln(a)[/itex] , i thought you only move the exponent out, but you moved both the base AND exponent.

    edit: Oh, I think i understand now. if you have a constant to the x power, like 52x then it becomes 52x ln5 ⋅2

    but what if it is a variable to a power of another variable, for example y5x

    is it going to be y5x ln y ⋅5 ?
     
    Last edited: Mar 1, 2015
  14. Mar 1, 2015 #13

    Mark44

    Staff: Mentor

    Here's how it works. As long as x > 0, you can write x as ##e^{ln(x)}##.
    So ##5^{2x} = (e^{ln(5)})^{2x}##
    Using one of the rules of exponents, the above is equal to ##e^{2x*ln(5)}##. This form is much easier to differentiate.
     
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