Question

What is the derivative of `Ln(x)/(1+ln(2x))`?

Answer 1

`d/dx lnx/(1+ln(2x))= (1+ln(2x) - lnx) / (x(1+ln(2x))^2)`

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`, or less formally, `(u/v)' = (v(du)-u(dv))/v^2`

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with `y=(lnx)/(1+ln(2x))` Then

`{ ("Let "u=lnx, => , (du)/dx=1/xx), ("And "v=1+ln(2x), =>, (dv)/dx=1/x ) :}`

`:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
`:. dy/dx = ( (1+ln(2x))(1/x) - (lnx)(1/x) ) / (1+ln(2x))^2`

`:. dy/dx = (1/x)(1+ln(2x) - lnx) / (1+ln(2x))^2`
`:. dy/dx = (1+ln(2x) - lnx) / (x(1+ln(2x))^2)`