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Math Help - Differentiation (ln)

  1. #1
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    Differentiation (ln)

    Hey guys, would really appreciate some help with this question. Its 2am where I am, and this is due tomorrow by 11:30am

    Heres the question:


    Differentiate the function:

    If somebody could give me a step by step worked out solution, I would be eternally grateful. Hope you guys can help
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  2. #2
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     y = \ln \left(\frac{x^{3}(6-x^{2})}{(x-1)(x^{4}+2)e^{x}} \right) .

    Then  \frac{dy}{dx} = \frac{(x-1)(x^{4}+2)e^{x}}{x^{3}(6-x^{2})}   \left[\frac{(x-1)(x^{4}+2)e^{x}(18x^{2}-5x^{4})-(x^{3}(6-x^{2})(xe^{x}(4x^{3}-4x^{2}+x^{4}+2))}{\left[(x-1)(x^{4}+2)e^{x} \right]^{2}} \right]

    Use quotient rule and and  \ln rule with chain rule.
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  3. #3
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    Quote Originally Posted by Archimedes View Post
    Hey guys, would really appreciate some help with this question. Its 2am where I am, and this is due tomorrow by 11:30am

    Heres the question:


    Differentiate the function:

    If somebody could give me a step by step worked out solution, I would be eternally grateful. Hope you guys can help
    It's too long for me to do right now, but notice that
    y = ln(f(x)) \implies y' = \frac{1}{f(x)} \cdot f'(x)

    So find the derivative of the function f(x) and put it over f(x). There should be a significant amount of cancellation before you are done.

    -Dan
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  4. #4
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    \ln(ab)=\ln a +\ln b,\,\ln\frac ab=\ln a-\ln b.

    Do the algebra before differentiating.
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  5. #5
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    Quote Originally Posted by particlejohn View Post
     y = \ln \left(\frac{x^{3}(6-x^{2})}{(x-1)(x^{4}+2)e^{x}} \right) .

    Then  \frac{dy}{dx} = \frac{(x-1)(x^{4}+2)e^{x}}{x^{3}(6-x^{2})}   \left[\frac{(x-1)(x^{4}+2)e^{x}(18x^{2}-5x^{4})-(x^{3}(6-x^{2})(xe^{x}(4x^{3}-4x^{2}+x^{4}+2))}{\left[(x-1)(x^{4}+2)e^{x} \right]^{2}} \right]

    Use quotient rule and and  \ln rule with chain rule
    .
    I really dont know how to do these rules with a sum this large, is there anyway you could show me how?
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  6. #6
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    Quote Originally Posted by Archimedes View Post
    I really dont know how to do these rules with a sum this large, is there anyway you could show me how?
    Have you tried using the very sensible approach suggested by Krizalid? Do you know how to apply the basic log rules to the given expression?

    Two other rules pertinent to doing this:

    \ln a^m = m \ln a

    \ln e^x = x (which is an application of the above rule)
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    \ln(ab)=\ln a +\ln b,\,\ln\frac ab=\ln a-\ln b.

    Do the algebra before differentiating.
    I got to where particlejohn got to (although when looking at his solution, I discovered I made some pretty basic errors in the process which would have distorted my overall solution) but Im stuck from there on!
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  8. #8
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    Quote Originally Posted by particlejohn View Post
     y = \ln \left(\frac{x^{3}(6-x^{2})}{(x-1)(x^{4}+2)e^{x}} \right) .

    Then  \frac{dy}{dx} = \frac{(x-1)(x^{4}+2)e^{x}}{x^{3}(6-x^{2})}   \left[\frac{(x-1)(x^{4}+2)e^{x}(18x^{2}-5x^{4})-(x^{3}(6-x^{2})(xe^{x}(4x^{3}-4x^{2}+x^{4}+2))}{\left[(x-1)(x^{4}+2)e^{x} \right]^{2}} \right]

    Use quotient rule and and  \ln rule with chain rule.
    particlejohn, i know you have been immensely helpful so far, but any chance you could post the following steps in this question like you have above?
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  9. #9
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    Quote Originally Posted by Archimedes View Post
    I got to where particlejohn got to (although when looking at his solution, I discovered I made some pretty basic errors in the process which would have distorted my overall solution) but Im stuck from there on!
    I will bet that the person who wrote this question expected it to be done by first simplifying the expression using log rules. I don't understand why you insist on banging your head against a much more difficult approach.
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  10. #10
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    Sorry, I guess its stubborness. I was doing it my way until I got stuck and was reluctant to start again, but also Im not very good at applying the log rules. Its almost 4am, my head (and eyes) are fried after a days work and study (thats what I get for failing this the first time round ) and I have to hand up this repeat paper in 7 hours. All these numbers are becoming a blur to me. Is there any chance of a step by step solution?
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Archimedes View Post
    Sorry, I guess its stubborness. I was doing it my way until I got stuck and was reluctant to start again, but also Im not very good at applying the log rules. Its almost 4am, my head (and eyes) are fried after a days work and study (thats what I get for failing this the first time round ) and I have to hand up this repeat paper in 7 hours. All these numbers are becoming a blur to me. Is there any chance of a step by step solution?
    following Krizalid's suggestion:

    the required rules are:

    (1) \log_a \frac xy = \log_a x - \log_a y

    (2) \log_a xy = \log_a x + \log_a y

    (3) \log_a a^x = x...in particular, note that \ln x \equiv \log_e x

    (4) \log_a x^n = n \log_a x



    Using these rules:

    y = \ln \Bigg[ \frac {x^3 (6 - x^2)}{(x - 1)(x^4 + 2)e^x} \Bigg]

    = \ln \Bigg[ x^3 (6 - x^2) \Bigg] - \ln \Bigg[ (x - 1)(x^4 + 2)e^x \Bigg] ........applied rule (1)

    = \ln x^3 + \ln (6 - x^2) - \ln (x - 1) - \ln (x^4 + 2) - \ln e^x ...........applied rule (2)

    = 3 \ln x + \ln (6 - x^2) - \ln (x - 1) - \ln (x^4 + 2) - x ....applied rules (4) and (3) to the first and last terms, respectfully

    Now continue by applying the rule topsquark gave you in post #3 to each term.
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  12. #12
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    Quote Originally Posted by Jhevon View Post
    Now continue by applying the rule topsquark gave you in post #3 to each term.
    I dont even know where to begin in applying that rule! But thanks for all the effort so far, Ill get some attempt marks anyway!
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  13. #13
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    Quote Originally Posted by Jhevon View Post
    following Krizalid's suggestion:

    the required rules are:

    (1) \log_a \frac xy = \log_a x - \log_a y

    (2) \log_a xy = \log_a x + \log_a y

    (3) \log_a a^x = x...in particular, note that \ln x \equiv \log_e x

    (4) \log_a x^n = n \log_a x



    Using these rules:

    y = \ln \Bigg[ \frac {x^3 (6 - x^2)}{(x - 1)(x^4 + 2)e^x} \Bigg]

    = \ln \Bigg[ x^3 (6 - x^2) \Bigg] - \ln \Bigg[ (x - 1)(x^4 + 2)e^x \Bigg] ........applied rule (1)

    = \ln x^3 + \ln (6 - x^2) - \ln (x - 1) - \ln (x^4 + 2) - \ln e^x ...........applied rule (2)

    = 3 \ln x + \ln (6 - x^2) - \ln (x - 1) - \ln (x^4 + 2) - x ....applied rules (4) and (3) to the first and last terms, respectfully

    Now continue by applying the rule topsquark gave you in post #3 to each term.
    Wait, I think Ive got the following....


    Would that be correct so far? And I just need to simplify that now......which I suck at Can anyone help?
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  14. #14
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    Bravo Bravo! You are correct.
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  15. #15
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    Quote Originally Posted by Chop Suey View Post
    Bravo Bravo! You are correct.
    Hehe thanks, the head is working a lot better after 6 hours sleep. Managed to get an extension until half 4 this evening (its 11:05am where I am), so Ive got some time now.... Still having trouble simplifying though - its always the easiest things u forget!
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