Logarithmic differentiation problem (homework)

Discussion in 'Math' started by jaygatsby, Mar 16, 2012.

  1. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    I have the following problem:

    f(x) = e^(4x-1) + 5^x

    I see that I need to use logarithmic differentiation. I've worked on this awhile and I am confused on two points:

    1) How do I do ln of the right side of the equation when there are two terms? ln each(ie. does ln distribute)? ln the whole thing and use algebraic properties of ln to turn it into ln(e^(4x-1)*5^x)?

    2) I don't know how to ln(e^(4x-1)). I haven't found a similar example in the textbook. In the answer to this question e^(4x-1) is a term. But from what I understand, ln(e^4x-1) should be (4x-1)e, as the whole point of using logarithmic differentiation is to get that exponent with an x out of there!

    Hellllllppppppppppp!

    Thanks,
    J
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    For number 1, you're going to slap your head when you see your oversight. :p

    The derivative of a sum is the sum of the derivatives. So. just break it into two pieces.

    f(x)=g(x)+h(x)

    where g(x)=exp(4x-1) and h(x)=5^x

    now f'(x)=g'(x)+h'(x)

    For number 2, remember that ln(exp(y))=y. This means that you can automatically get x out of the exponent once you take the logarithm of exp(4x-1), or just directly find the derivative of exp(4x-1) since it is an exponential function that you know how to differentiate. Don't worry about what the answer says and try to compare. Just work it out slowly and it will all fall into place.
     
    Last edited: Mar 16, 2012
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  3. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    Thank you. I am still having problems, it seems.

    <br />
f(x) = e^{4x-1} + 5^x<br />

    Find f'(x).

    I need to use logarithmic differentiation since 5^x needs it, I need to put it on both sides of the equation, to maintain equality. Then I'll simplify using what I know about ln. For instance, ln(e^anything) = anything.

    <br />
y = e^{4x-1} + 5^x<br />
ln(y) = ln(e^{4x-1}) + ln(5^x)<br />
ln(y) = (4x-1)+x ln(5)<br />

    I differentiate the first term, and use the chain rule on the second term. Also, ln(constant) is differentiated as 1/constant

    <br />
\frac{1}{y} \frac{dy}{dx} = 4+[x\frac{1}{5}+ln(5)]<br />

    :(
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    No, that's not quite right. Logarithm of a sum is NOT equal to the sum of the logarithms.

    You missed my point about the fact that a derivative of a sum is equal to the sum of the derivatives. Do you know this basic rule?

    You are trying to find f'(x), right?

    So, break it into pieces

    f(x)=g(x)+h(x) and f'(x) = g'(x)+h'(x)

    The g(x) is the exponential piece and you can do this without resorting to the logarithmic differentiation technique (although, you are free to do it that way if you choose).

    This leaves the h'(x) to be calculated separately using logarithmic differentiation.
     
    Last edited: Mar 16, 2012
  5. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    I understand that point, but I don't understand how I can just take the ln of the h function. Won't doing something to one term on the right hand side of the = change the equality? Don't I have to take ln() of both sides for the equality to remain?

    Thank you
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    Yes, I understand your concern, but the way to do it is to calculate the derivatives of each term separately. It doesn't really matter how you determine h' and g' because once you add them together, you have f'.

    So think about solving two problems separately, and then add the results together. As separate terms you are free to take the log of g or h and then take the derivative, which then leads to the individual derivatives g' and h'.
     
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  7. jaygatsby

    Thread Starter New Member

    Nov 23, 2011
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    This works thanks. Why do all the examples in my book, and on mathtv.com, show logarithmic differentiation by taking the ln of both sides, then arranging so that dy/dx is on one side? If ln diff can be done with just one term, I assume it must apply for the case where there's just one term in the equality, so taking the ln of y wouldn't be necessary at all :\
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    Well examples are often shown to demonstrate the method, then we are expected to integrate that method into other methods to find the answers to more general problems.

    I prefer to use a slightly different variation to the logarithmic differentiation method. If you have this example y=g+h, you can write it as follows

    y=exp(ln(g))+exp(ln(h))

    The above is true because exp and ln are inverse functions.

    The derivative of exp is easy and the log inside allows the simplifications in the exact same way. So, this form of the method works for any number of terms, but obviously you would only use it on the relavent terms.
     
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