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

1 Answer
Steve M
Nov 8, 2016

# 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) #

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