
Bradford T. answered 10/11/24
Retired Engineer / Upper level math instructor
y=(ln(7x))ln(2x)
Take log of both sides
ln(y)=ln((ln(7x))ln(2x))
ln(y)=ln(2x) ln(ln(7x)) log identity ln(ab) = bln(a)
Take derivative
y'/y = (1/(2x))(2)ln(ln(7x)) + ln(2x)(1/ln(7x))(1/(7x))(7)
y' = y[ln(ln(2x))/x +ln(2x)(1/ln(7x))(1/x)]
where y = (ln(7x))ln(2x)