anonymous
  • anonymous
the derivative of x^(-2x) is -2x^(-2x) (log(x)+1) why is there that +1?
Mathematics
chestercat
  • chestercat
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mathteacher1729
  • mathteacher1729
See image. 1) Go to http://www.wolframalpha.com 2) type "derivative x^(-2x) Hope this helps. :)
anonymous
  • anonymous
\[x^{-2x}\] take the log get \[-2x\ln(x)\] take the derivative using product rule get \[-2x\frac{1}{x} -2ln(x)\]
anonymous
  • anonymous
now factor out the 2 get \[2(\ln(x)-1)\]

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anonymous
  • anonymous
of course the x's in the first term cancel. then multiply by the original function to get your answer: \[x^{-2x}(2(ln(x)-1)=2\times x^{-2x}(ln(x)-1)\]
anonymous
  • anonymous
i made a mistake is see. start with \[-2-2ln(x)\] gives \[-2(ln(x)+1)\]
anonymous
  • anonymous
sorry about that.
anonymous
  • anonymous
but that is where the -2 comes from
anonymous
  • anonymous
you can do this because ln(x) '=1/x so if you take the log and at the end you multiply by the function than you get the derrivate
anonymous
  • anonymous
this is a question :)
anonymous
  • anonymous
oh yes. if you want to take the derivative of a function with the variable in the exponent the easiest thing to do is first take the log.
anonymous
  • anonymous
I have never seen this before, but it is a clever idea
anonymous
  • anonymous
you know that the derivative of \[ln(f(x))\] is \[\frac{f'(x)}{f(x)}\]
anonymous
  • anonymous
yep, I see
anonymous
  • anonymous
which means you take the log, take the derivative, and then multiply by original function.
anonymous
  • anonymous
yes yes, cool technique
anonymous
  • anonymous
you can also view it this way: \[x^{2x}=e^{-2x\ln(x)}\]
anonymous
  • anonymous
then take the derivative using the chain rule. the derivative of e to the something is just e to the something times the derivative of something. you will see that the work is identical
anonymous
  • anonymous
thanks, I see why this helps
anonymous
  • anonymous
in any case all the work is taking the derivative of the log of that thing, either way. hope this helps

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