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Calculating a deriv using logarithmic differentiation

  1. Feb 22, 2008 #1
    Hello,This is my first crack at using log differentiation, but I can't seem to get too far with it...

    Use logarithmic differentiation to calculate the derivative for the following function:

    [tex] y = \sqrt{x}e^{x^{2}} (x^{2} + 1)^{10} [/tex]

    [tex] lny = \frac{1}{2}lnx * x^{2}lne * 10ln(x^{2} + 1) [/tex]

    (I'm not sure what to do with the deriv of lne^{x^{2}})

    [tex] lny = \frac{1}{2} * \frac{1}{x} * [deriv of lne^{x^{2}}] * 10 *\frac{2x}{x^{2} + 1} [/tex]

    [tex] lny = \frac{20x}{2x(x^{2} + 10} * [deriv of lne^{x^{2}}] [/tex]
  2. jcsd
  3. Feb 22, 2008 #2


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    Homework Helper

    I am assuming that you use '*' for multiplication.

    You can notice that, taking derivative of summation is 'easier' than those of multiplication.

    Say: (u + v + w)' = u' + v' + w' looks far easier compared with (uvw)' = u'vw + uv'w + uvw', right?

    And using log function, one can change multiplication, into summation. Hence, making it more comfortable when dealing with a bunch of multiplication.

    Nah, you are wrong from the first line :(, you should note that:

    ln(xy) = ln(x) + ln(y), it's not ln(xy) = ln(x) * ln(y) as you've written.

    I'll re-do it for you. Let's begin:

    [tex]y = \sqrt{x} e ^ {x ^ 2} (x ^ 2 + 1) ^ {10}[/tex]

    Take logs of both sides yields:

    [tex]\Rightarrow \ln( y ) = \ln \left( \sqrt{x} e ^ {x ^ 2} (x ^ 2 + 1) ^ {10} \right)[/tex]

    Use the formula: ln(xy) = ln(x) + ln(y), and ln(ab) = b ln(a) to expand all the terms on the RHS:

    [tex]\Rightarrow \ln( y ) = \frac{1}{2} \ln (x) + x ^ 2 \ln (e) + 10 \ln (x ^ 2 + 1)[/tex]

    ln(e) = 1, so, let's simplify it a bit:

    [tex]\Rightarrow \ln( y ) = \frac{1}{2} \ln (x) + x ^ 2 + 10 \ln (x ^ 2 + 1)[/tex]

    If the two functions (say, f(x), and g(x)) are the same (f(x) = g(x), for all x), then so are their derivatives (i.e, f'(x) = g'(x), for all x).

    So, taking the derivatives of both sides with respect to x (not just one side as you did), we have:

    [tex]\Rightarrow \frac{y'_x}{y} = \left( \frac{1}{2} \ln (x) + x ^ 2 + 10 \ln (x ^ 2 + 1) \right)'_x[/tex]

    Now, what you have to do is to take the derivatives of the RHS, then multiply y over, and arrive at the desired answer.

    Hope that you can go from here. :)


    Btw, you should put a '\' in front of a function in LaTeX to make it looks un-italics.

    Compare the two:

    \ln x returns: [tex]\ln x[/tex]

    whereas ln x returns: [tex]ln x[/tex]

    The first one looks somewhat better, eh? :)

    And to insert texts in LaTeX, we use the function \mbox{your text here}.

    [tex]\mbox{LaTeX is wonderful :D}[/tex] is way better than [tex]LaTeX is wonderful :D[/tex]. See? :)

    Hint: Click on LaTeX images to see ther codes..
    Last edited: Feb 23, 2008
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