How do you differentiate $f(x)=x^-3ln(1/x)+4ln(x^4)$ ?

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Answer 1 · 2015-11-04T04:55:05

Because

$ln(1/x)=-lnx$ and $lnx^4=4*lnx$ we have that

$f(x)=-x^-3*lnx+16lnx$ hence its derivative is

$df(x)/dx=3*x^-4*lnx-x^-3*(1/x)+16*(1/x)=> df(x)/dx=3*x^-4*lnx-x^-4+16*x^-1$

How do you differentiate f(x)=x^-3ln(1/x)+4ln(x^4)f(x)=x^-3ln(1/x)+4ln(x^4)?