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Mathematics 18 Online
OpenStudy (anonymous):

derivative of lxl?

OpenStudy (barrycarter):

http://fwd4.me/17pC

OpenStudy (anonymous):

thanks a lot.it made me available to find additional information of it?

OpenStudy (turingtest):

the key is the definition:\[|x|=\sqrt{x^2}\]using the chain rule we get\[\frac12(x^2)^{-1/2}(2x)=\frac x{(x^2)^{1/2}}={x\over\sqrt {x^2}}\]

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