How do you differentiate $f(x)=xln(1/x)-lnx/x$ ?

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Answer 1 · 2016-01-03T15:02:49

It is $f'(x)=ln(1/x)-1-(1-lnx)/x^2$

The derivative is

$d(f(x))/dx=x'*ln(1/x)+x*ln(1/x)'-((lnx)'x-x'*lnx)/(x^2)=> d(f(x))/dx=ln(1/x)+x*((1/x)'/(1/x))-(1-lnx)/x^2=> d(f(x))/dx=ln(1/x)-1-(1-lnx)/x^2$

Remember that

$d(lnx)/dx=1/x$

How do you differentiate f(x)=xln(1/x)-lnx/x f(x)=xln(1/x)-lnx/x ?