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Computing derivative (basic calculus question)

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Compute Derivative
    y = xx + sin(x)
    2. Relevant equations


    3. The attempt at a solution
    since I have x in the exponent (x^x), I multiply both sides by ln:

    ln y = ln xx + ln sin(x)

    the x in the exponent comes out into the front, right?

    y'/y = x ln x + ln sin (x)

    using product rule for xlnx:

    y'/y= ((1⋅ln x) + (x) (1/x)) + 1/sin(x) ⋅ cos(x)

    y'/y = [(ln x +1) + cos(x)/sin(x)]

    multiplying both sides by y to get y' only

    y' = [(ln x +1) + cos(x)/sin(x)] ⋅ xx + sin(x)



    is this correct? or should the ln xx = Xx ⋅ ln x ???
     
  2. jcsd
  3. Feb 28, 2015 #2

    SteamKing

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    In order to take a logarithm of a number, you do not multiply that number by 'ln', just like if you take the sine of an angle, you do not multiply the angle by 'sin'.

    The logarithm is a function, just like the sine is a function. Before you dive into calculus, you should learn what functions are.
     
  4. Feb 28, 2015 #3

    SammyS

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    ln(a + b) ≠ ln(a) + ln(b) .

    By the way, that is not multiplying ln times (a + b) .

    Hint: Find the derivative of xx separately, by using the natural log (ln) and implicit differentiation. Then include the sine term.
     
  5. Feb 28, 2015 #4

    LCKurtz

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    In addition to SteamKing's observations, since you already know how to differentiate ##\sin x##, just do ##x^x## separately by your method and add it to the derivative of ##\sin x##. It will help you avoid the mistakes you made.

    [Edit] Didn't see Sammy's reply.
     
  6. Feb 28, 2015 #5
    i saw 2 videos on youtube, apparently you can get the derivative of xx using e and another way without using e, is this correct?

    both come out to be xx (lnx+1)
     
  7. Feb 28, 2015 #6

    SammyS

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    You can express xx in an alternative form using e. This does make it relatively easy to find the derivative .

    ##\displaystyle x^x = e^{\ln(x^x)}=e^{\,x\,\ln(x)}##
     
  8. Feb 28, 2015 #7
    ok, I got the new answer now. is it:

    y' = (Xx (ln x + 1) + cos(x)) / xx + sin(x)
     
  9. Feb 28, 2015 #8

    SammyS

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    The trig functions should be completely separate from the xx .
     
  10. Feb 28, 2015 #9
    what do you mean by that?
     
  11. Feb 28, 2015 #10

    SammyS

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    You have cos(x) divided by xx .
     
  12. Feb 28, 2015 #11
    any hint on how to get rid of the trig functions? I have no idea what to do at this point.
     
  13. Mar 1, 2015 #12

    SammyS

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    You don't get rid of them. They are just not in the same term as xx .

    ##\left(f(x)+g(x)\right)'=f'(x)+g'(x) ##
     
  14. Mar 1, 2015 #13
    wow, I forgot to multiply both sides by y (the original equation)

    then after I do all that, the base and y cancel out and I'm left with X^x (lnx+1) + cos (x))

    correct?
     
  15. Mar 1, 2015 #14

    HallsofIvy

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    i
    In addition to what others have said, this is wrong. You have differentiated on the left side but not yet on the right- they are NOT equal!

     
  16. Mar 1, 2015 #15

    SammyS

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    That is the correct derivative.

    Once you find the derivative of xx , you shouldn't need to multiply anything by y .
     
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