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anonymous
 5 years ago
the derivative of x^(2x) is 2x^(2x) (log(x)+1)
why is there that +1?
anonymous
 5 years ago
the derivative of x^(2x) is 2x^(2x) (log(x)+1) why is there that +1?

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mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.1See image. 1) Go to http://www.wolframalpha.com 2) type "derivative x^(2x) Hope this helps. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^{2x}\] take the log get \[2x\ln(x)\] take the derivative using product rule get \[2x\frac{1}{x} 2ln(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now factor out the 2 get \[2(\ln(x)1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0of 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
 5 years ago
Best ResponseYou've already chosen the best response.0i made a mistake is see. start with \[22ln(x)\] gives \[2(ln(x)+1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but that is where the 2 comes from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0this is a question :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0I have never seen this before, but it is a clever idea

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know that the derivative of \[ln(f(x))\] is \[\frac{f'(x)}{f(x)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which means you take the log, take the derivative, and then multiply by original function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes yes, cool technique

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can also view it this way: \[x^{2x}=e^{2x\ln(x)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.0thanks, I see why this helps

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in any case all the work is taking the derivative of the log of that thing, either way. hope this helps
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