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Homework Help: Differentiate this expression with respect to x

  1. Apr 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Differentiate the following with respect to x leaving in the simplest form
    \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right]

    2. Relevant equations
    The chain rule
    \frac{d}{{dx}}\left[ {\left( {f\left( x \right)} \right)^n } \right] = nf\left( x \right)f'\left( x \right)
    The product rule
    \frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)g'\left( x \right) + f'\left( x \right)g\left( x \right)
    The natural log rule
    \frac{d}{{dx}}\left[ {\ln f\left( x \right)} \right] = \frac{{f'\left( x \right)}}{{f\left( x \right)}}

    3. The attempt at a solution

    I sort of understand how to differentiate the following, however i cannot get it into the simplest form. What would the best way to differentiate this functions and others which have a similar format?
    many thanks for all suggestions and help
  2. jcsd
  3. Apr 15, 2007 #2


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    Well use the three rules you quote. Your expression can be written as ln[f(x)g(x)] where [itex]f(x)=(1-\sqrt{x})^2, \hspace{1cm} g(x)=(1+\sqrt{x})^3[/itex]. Now, first apply the log rule to give [tex]\frac{1}{f(x)g(x)}\cdot\frac{d}{dx}[f(x)g(x)][/tex]. Then apply the product rule, followed by the chain rule. This is the simplest way, as far as I can tell; of course, once you have differentiated, you will need to simplify the expression you obtain.

    If you show some work, it'll be easier to help!

    [as an aside, this isn't precalculus maths is it?]
    Last edited: Apr 15, 2007
  4. Apr 15, 2007 #3
    simplify f(x)g(x) using simple algebra properties first then only apply some log rule like log ab = log a + log b
    i'm not too sure if d/dx log ab = d/dx log a + d/dx log b, but you may give it a try
  5. Apr 15, 2007 #4
    thank you for your reply cristo
    this problem was given to me in my early stages of calculus, so i though of posting it here, as it what i have considered a fundamental stage, similar to pre-calculus
    here is my working for the problem. i first simplified the expression before attempting to differentiate it.
    \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right] = \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right]} \right] \\
    \frac{1}{{f\left( x \right)g\left( x \right)}}.\frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] \\
    \frac{d}{{dx}}\left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right] = \left( {1 + \sqrt x } \right)\left( {2x} \right) + \left( {x - 1} \right)^2 \left( {\frac{1}{{2\sqrt x }}} \right) \\
    \frac{d}{{dx}}\left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right] = \frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x }} \\
    \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right] = \frac{{\frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x }}}}{{\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 }} = \frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x \left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 }} = \frac{{4\sqrt x + 4x + x^2 - 2x + 1}}{{\left( {2\sqrt x + 2x} \right)\left( {x^2 - 2x + 1} \right)}} \\

    i am a little stuck on how to further simplify this down, as my other working seems to be going around in circles. any help is greatly appreciated.
  6. Apr 15, 2007 #5
    chickens thank you for your post, i will try that
  7. Apr 15, 2007 #6


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    It's always a good idea to simplify as much as possible before differentiating:
    [tex]\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]= 2 ln(1- \sqrt{x})+ 3ln(1+ \sqrt{x})[/tex]
  8. Apr 15, 2007 #7
    thanks HallsofIvy
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