How do you differentiate $\frac{(e^{x}) (ln x)}{x}$ $[(e^x)(lnx)]/x$ ?

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Answer 1 · 2016-08-19T14:23:39

$\frac{e^{x}}{x^{2}} (1 + (x - 1) ln x)$ $e^x/x^2(1+(x-1)lnx)$

$f (x) = \frac{e^{x} ln x}{x}$ $f(x) = (e^xlnx)/x$

$f'(x) = (x(e^x*1/x+e^xlnx)-e^xlnx*1)/x^2$
(Quotient rule and Produce rule)

$f'(x) = (e^x+xe^xlnx-e^xlnx)/x^2$

$= e^x/x^2(1+(x-1)lnx)$

How do you differentiate [(e^x)(lnx)]/x (ex)(lnx)x [(e^x)(lnx)]/x ?