Question

What is the derivative of y = log(2x)? Help!?

Answer 1

If the latter resented is natural log, the answer is `1/x`

Explanation:

The derivative `d/(dx) ln (u (x)) = ((du)/(dx))/(u (x))`

If the log in the problem is natural log rather than common log, this derivative appliea. Given that `u (x) = 2x, u'(x)=2`, and thus...

`d/(dx)ln (2x) = 2/(2x) = 1/x`

If instead we are dealing with common log, then because `log_a (x)= ln (x)/(ln a)`, we would divide our result by `ln a`, ln 10 in this case

Answer 2

Looking at the wording in your question there is a trap!

`dy/dx=1/(xln(10))`

Explanation:

The format of `log(2x)` is normally interpreted as `log_10(x)`

`color(brown)("Assumption: The question really is talking about log to base 10.")`

It is known that `d/(du)(ln(u))=1/u` so lets take advantage of this by

converting `log_10(2x)` to natural logs

Set `u=2x" "=>" "(du)/dx=2`

Set `y=log_10(u)`

Then we have:

`y=ln(u)/(ln(10))`

But `1/(ln(10))` is a constant so we have

`dy/(du) =1/(ln(10))xx 1/u`

But `dy/(du)xx(du)/(dx)=dy/dx` so we have:

`dy/dx=1/(ln(10))xx1/(2x)xx2`

`dy/dx=1/(xln(10))`