What is the derivative of $y=log_10x/x$ ?

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Answer 1 · 2016-10-09T06:48:10

$(dy)/(dx)=log_10e((1-lnx)/x^2)=0.4343((1-lnx))/x^2$

$y=log_10x/x$ can be written as $y=log_10 exxlnx/x$

or $y=0.4343xxlnx/x$ and using quotient rule

Hence $(dy)/(dx)=0.4343xx(x xx1/x-lnx xx1)/x^2$

= $0.4343((1-lnx))/x^2$

What is the derivative of y=log_10x/xy=log_10x/x?