Question

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

Answer 1

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

Explanation:

`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`