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Need help with a Differentiation Question

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Differentiate f(x) if f(x)=xxln(3x-6)

    2. Relevant equations

    None

    3. The attempt at a solution

    The problem in f(x) is the xx so what I did was let a=xx and differentiated a with respect to x. I ended up with da/dx= (ln(x)+1)xx. Afterwards, I modified f(x) by substituting a for xx and ended up with f(x)=aln(3x-6) and differentiated that equation which gives me: f'(x)=(ln(x)+1)(ln(3x-6))(xx)+(xx/(x-2)).

    Is this correct?
     
  2. jcsd
  3. Nov 6, 2009 #2
    Hey Sky.Anthony :D,

    Yeh that looks good to me, now Im not sure how you did

    [tex]\frac{d}{dx}(x^x)[/tex]​

    Im quessing as you let a=xx you did ln(a) = xln(x) and differentiated implicitly. However another way of doing this is considering

    [tex]x^x = e^{ln(x^x)} = e^{xlnx}[/tex]​

    although you may have done it that way anyway, and of course one can come to this result from the previous, in fact that's probably how its supposed to be done, but just in case you didn't, its a lovely little gem I feel :D
     
  4. Nov 6, 2009 #3
    Hey, thanks for your reply.

    Once I got a=xx, I differentiated implicitly...
    ln(a) = x ln(x)
    da/dx 1/a = [ ln(x)+1 ]
    da/dx = xx (ln(x)+1)

    And then in the original function f(x)= a ln (3x-6), once I had to differentiate a, I just substituted in da/dx. I'm just curious, when you said that my answer looks good, did you mean that I am working through the problem correctly or did you mean that my answer was correct? The reason I'm asking is that the answer to this question will determine whether I get 50% for this part in my assignment or 100% :)
     
  5. Nov 6, 2009 #4
    Hey Sky.Anthony,

    Sorry, i did mean the answer you worked out was correct, the way you did it is perfectly fine, using a substitution certainly makes the differentiating the function easier somewhat as one doesn't have to deal with long complex expression, but without using a substitution as you did, you could do:

    [tex]\begin{array}{rcl}
    \displaystyle \frac{d}{dx}\left(x^{x}ln(3x-6)\right)&=&\displaystyle \frac{d}{dx}(e^{xlnx}ln(3x-6))\\\\
    &=&\displaystyle e^{xlnx}\frac{d}{dx}\left(ln(3x-6)\right) + ln(3x-6)\frac{d}{dx}(e^{xlnx})\\\\
    &=&\displaystyle e^{xlnx}\frac{3}{3x-6} + ln(3x-6)\frac{d}{d xlnx}(e^{xlnx})\frac{d}{dx}(xlnx)\\\\
    &=&\displaystyle e^{xlnx}\frac{1}{x-2} + e^{xlnx}ln(3x-6)\left(x\frac{d}{dx}(lnx) + lnx\frac{d}{dx}(x)\right)\\\\
    &=&\displaystyle \frac{e^{xlnx}}{x-2} + e^{xlnx}ln(3x-6)\left(x\left(\frac{1}{x}\right) + lnx\right)\\\\
    &=&\displaystyle \frac{x^x}{x-2} + x^x(1 + lnx)ln(3x-6)
    \end{array}
    [/tex]​

    which is exactly what you have got :D, so hopefully you can see in that the application of the various rules; chain and product, but as long as you understand how you got to your answer that's the important thing, cheers Sky.Anthony :D
     
    Last edited: Nov 7, 2009
  6. Nov 6, 2009 #5
    Awesome! Thanks so much :)
     
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