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    nccaitlin91's Avatar
    nccaitlin91 Posts: 26, Reputation: 1
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    #1

    May 5, 2015, 12:42 PM
    derivative of y=ln(3xcosx)
    Hi, one of my problems for technical calculus is:

    Find the derivative of Y=ln(3xcosx)

    My logarithmic examples have not included trig functions. I am thinking I will have to set this up as product rule somehow and then take the natural log, correct?

    y1= (3x) [cosx] + (cos x) [3x]
    y1= 3cosx + (-sinx) (3)
    y1= 3cosx - 3sinx
    if that is correct, how do I take the natural log of trig functions?

    Thanks so much. I have tried using my notes and book but none of the examples were set up like this.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    May 5, 2015, 12:48 PM
    If then the derivative is:



    For this problem you have . Try it again, and post back with your results.
    nccaitlin91's Avatar
    nccaitlin91 Posts: 26, Reputation: 1
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    #3

    May 5, 2015, 12:55 PM
    Quote Originally Posted by ebaines View Post
    If then the derivative is:



    For this problem you have . Try it again, and post back with your results.

    If I am understanding you (which I might not be because I am terrible with math)

    y1= 1/ [(3x)(-sinx)]

    To get rid of -sin, change to

    y1= (1/-3xsinx)
    or -(1/3xsinx)
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    May 5, 2015, 01:35 PM
    No. f(x) = 3x cos(x), So f'(x) = 3x(-sin(x))+3cos(x).

    Let me show you a different but similar problem, so you can get the idea: find the derivative of . This is of the form where. Recall that the derivative of tan(x) is 1/cos^2(x), so:




    You can use this same technique for your problem.

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