How do you differentiate $\frac{e^{x}}{ln x}$ $e^x/ln x$ ?

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Answer 1 · 2016-03-15T07:21:06

$y'=(e^x(xlnx + 1))/(xln^2x)$

$y=f/g$

$y'=(f'*g-f*g')/g^2$

$y=e^x/lnx => f=e^x, g=lnx$

$f'=e^x, g'=1/x$

$y'=(e^xlnx - e^x/x)/ln^2x$

$y'=(e^x(xlnx - 1))/(xln^2x)$

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