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

Question

29820 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-25T19:05:34

We can use the chain rule here, naming #u=2x# and remembering that the chain rule states that

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

So, now, for our function #ln(u)#:

#(dy)/(du)=1/u#

And for the other part:

#(du)/(dx)=2#

Now, aggregating them:

#(dy)/(dx)=1/u*2=1/(cancel(2)x)*cancel2=color(green)(1/x)#