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Find dy if y=(e^x)logx
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go get'em cowboy
i assume this means \[\frac{d}{dx}e^x \ln(x)\] but i could be wrong
it says log, not natural log, so I assume its log of ten, so ln...
Use product rule \[dy = \frac{d}{dx}e^{x}*\log x + e^{x}*\frac{d}{dx}\log x\] \[dy = e^{x}\log x + \frac{e^{x}}{x} \]
wait i assumed natural log
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if it is \[\log_{10}(x)\] the the derivative is \[\frac{\ln(10)}{x}\]
yep
right. ok thanks.
your welcome :)
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