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Math Help - Logarithmic Differentiation with binomial products

  1. #1
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    Logarithmic Differentiation with binomial products

    Hi,

    I've been asked to differentiate y = 5x^2tan(6x) logarithmically.

    I took the following steps:

    y = 5x^2tan(6x)

     ln(y) = ln(5x^2tan6x) = (2ln5x)(lntan(6x))

    f(x) = 2ln5x
    g(x) = lntan(6x)
    f'(x) = \frac {2}{x}
    g'(x) = \frac {1}{6Sec^26x}

    \frac {1}{y} \frac{dy}{dx} = (2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6x  ))

    \frac {dy}{dx} = y((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(  6x)))

    \frac{dy}{dx} = 5x^2tan(6x)  ((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6  x)))

    This is my first attempt at differentiating binomial products logarithmically, can anyone please look at these steps and tell me if and how I've gone wrong?

    Thanks.

    (Please spare any factorising, I don't understand it very well right now)
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  2. #2
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    Quote Originally Posted by ashleysmithd View Post
    y = 5x^2tan(6x)

     ln(y) = ln(5x^2tan6x) = (2ln5x)(lntan(6x))
    The log of a product is a sum of two logs - not a product of 2 logs.

    Quote Originally Posted by ashleysmithd View Post
    f(x) = 2ln5x
    g(x) = lntan(6x)
    f'(x) = \frac {2}{x}
    g'(x) = \frac {1}{6Sec^26x}
    So you don't want the product rule, though you do want to differentiate f and g. Your f is fine, but g dodgy.

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).



    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; January 2nd 2011 at 10:17 AM. Reason: img address
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  3. #3
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    \ln(y) = \ln(5) + 2\ln(x) + \ln(\tan(6x))

    The first two terms are simple enough to differentiate but you will need to use the chain rule twice on the last term

    Let u = tan(6x) \implies \dfrac{du}{dx} = 6\sec^2(6x) and also gives y = \ln(\tan(6x)) = \ln(u) and by the chain rule \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}

    (note that y here is used as an example to demonstrate the chain rule)
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  4. #4
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    I think I see what you mean, g(x) was a function of a function and needed the chain rule to differentiate it into: (\frac{1}{tan(6x)}(6sec^26x)) ?

    I'm not quite what you meant by using the chain rule twice, I've equated it to this so far:
    \frac{dy}{dx} = 5x^2tan6x ((2ln5x)(\frac{1}{tan(6x)}(6sec^26x))+(lntan6x)(\f  rac{2}{x}))

    Is this on the right lines?
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  5. #5
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    Quote Originally Posted by ashleysmithd View Post
    y = 5x^2tan(6x) logarithmically.
    Why make it so complicated?
    The derivative is simply: 10x\tan(6x)+30x^2\sec^2(6x).
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  6. #6
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    I completely understand it can be factorised down, but I really don't understand factorising well yet so I'd rather work long hand and just get the rules of differentiation right first.
    That said, all be it unnecessarily long, is the solution I posted above correct?
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  7. #7
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    It is a simple product: 5x^2 times \tan(6x).

    Use the product rule. Anything else is a waste of time and effort.
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  8. #8
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    I would do but the point of the exercise I've given is to learn how to differentiate logarithmically, and show working, so I can't use the product rule on its own annoyingly.
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  9. #9
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    Quote Originally Posted by ashleysmithd View Post
    I think I see what you mean, g(x) was a function of a function and needed the chain rule to differentiate it into: (\frac{1}{tan(6x)}(6sec^26x)) ?
    Good.

    I'm not quite what you meant by using the chain rule twice, I've equated it to this so far:
    \frac{dy}{dx} = 5x^2tan6x ((2ln5x)(\frac{1}{tan(6x)}(6sec^26x))+(lntan6x)(\f  rac{2}{x}))

    Is this on the right lines?
    No, it looks like you're still trying to involve the product rule even though the one advantage of doing it logarithmically (although, as Plato says, this example is not remotely a good case for that treatment) is that you got rid of the product so you don't need that rule.

    Look at the bottom row of my diagram and multiply both sides by y, perhaps?
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  10. #10
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    I think the part where I'm getting confused is how
    ln(y) = ln(5x^2tan6x)
    becomes
    ln(y) = 2ln5x + lntan6x

    I don't understand why these functions are being added and not multiplied? Why does it become a sum, and not remain a product, is it to do with log properties?
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  11. #11
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    Quote Originally Posted by ashleysmithd View Post
    is it to do with log properties?
    Yes - revise them.

    \ln(ab)\ =\ \ln(a) + \ln(b)\ \ \

    from

    a = e^x and b = e^y

    and

    ab = e^{x+y}
    Last edited by tom@ballooncalculus; January 3rd 2011 at 12:48 PM.
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  12. #12
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    Quote Originally Posted by ashleysmithd View Post
    I think the part where I'm getting confused is how
    ln(y) = ln(5x^2tan6x)
    becomes
    ln(y) = 2ln5x + lntan6x
    It doesn't - you can only take a power out when each term is raised to the power. You moved the 2 to the front but only the x was squared, not the 5

    \ln(y) = \ln(5x^2 \tan(6x)) = \ln(5x^2) + \ln(\tan(6x))

    You need to use the addition rule on \ln(5x^2) before the power rule can be done on x^2 -> \ln(5x^2) = \ln(5) + \ln(x^2) = \ln(5) + 2\ln(x)

    Hence \ln(y) = \ln(5) + 2\ln(x) + \ln(\tan(6x))

    Use the chain rule (explained in posts 2 and 3) to differentiate \ln(\tan(6x))
    Last edited by e^(i*pi); January 3rd 2011 at 12:56 PM. Reason: forgot a + sign
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  13. #13
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    I you want to learn Logarithmic Differentiation then do right.
    Suppose that each of g,~h,~j,~\&~k is a differentiable function and f(x)=g(x)\cdot h(x)\cdot j(x)\cdot k(x)\cdot .
    Then the derivative is f'(x) = f(x)\left[ {\frac{{h'(x)}}{{h(x)}} + \frac{{g'(x)}}{{g(x)}} + \frac{{j'(x)}}{{j(x)}} + \frac{{k'(x)}}{{k(x)}}} \right].
    Not that can be done for any numbers of factors.

    So the derivative of 5x^2\tan(6x) is simply 5x^2 \tan (5x)\left[ {\frac{2}{{x^2 }} + \frac{{6\cos (6x)}}{{\sin (6x)}} + \frac{{6\sin (6x)}}{{\cos (6x)}}} \right].

    But this is a horrible example to teach that concept.
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  14. #14
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    Ok thanks, I think I just got caught out on the log rules.
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