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

Question

5750 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-29T19:41:57

Let's start by finding the derivative of #ln(2x)#.

Let #y = lnu# and #u = 2x#.

Then #y' = 1/u# and #u' = 2#.

#dy/dx = 1/u xx 2 = 2/(2x) = 1/x#

We can now use the quotient rule to differentiate the entire function.

#dy/dx = (1/x xx x - ln(2x) xx 1)/(x)^2#

#dy/dx = (1-ln(2x))/x^2#

Hopefully this helps!

Answer 2 · 2016-10-29T19:50:31

The derivative is #=(1-ln(2x))/x^2#

This is the derivative of a quotient
#(u/v)'=(u'v-uv')/v^2#
so here
#u=ln(2x)# #=>##u'=2/(2x)=1/x#

#v=x##=>##v'=1#

So, #(ln(2x)/x)'=(x*1/x-1*ln(2x))/x^2#

#=(1-ln(2x))/x^2#