# Thread: logs and expontential functions

1. ## logs and expontential functions

I am not sure if I am getting this or not. Would you please check a couple of problems before I go any further.

f(x) = ln2x
f'(x) =1/u (du/dx 1/2x)(d/dx2x)
f'(x) = 2/2x

and
f(x) = x ln x
f'(x) = x lnx^3/2
f'(x) = x(1/2lnx)

2. Originally Posted by startingover
I am not sure if I am getting this or not. Would you please check a couple of problems before I go any further.

f(x) = ln2x
f'(x) =1/u (du/dx 1/2x)(d/dx2x)
f'(x) = 2/2x

and
f(x) = x ln x
f'(x) = x lnx^3/2
f'(x) = x(1/2lnx)
The first is correct, however the second isn't.

The second problem involves product rule: (u*v)' = u'*v + u*v'
f(x) = x lnx
f'(x) = (x)'lnx + x(lnx)'
f'(x) = (1)lnx + x(1/x)
f'(x) = lnx + 1

3. Originally Posted by startingover
I am not sure if I am getting this or not. Would you please check a couple of problems before I go any further.

f(x) = ln2x
f'(x) =1/u (du/dx 1/2x)(d/dx2x)
f'(x) = 2/2x

and
f(x) = x ln x
f'(x) = x lnx^3/2
f'(x) = x(1/2lnx)
Just for clarity, I would write the first problem this way:

f(x) = ln(2x)

Let u = 2x <--> u' = 2

f(x) = ln(u)
f'(x) = (ln(u))'*(u)'
f'(x) = (1/u)*2
f'(x) = 2/(2x) = 1/x

However, you don't always need to do a substitution for a problem like this. Try to get used to doing this problem without doing substitutions because they can be hard to keep track of in harder problems (and they take more work to write).

4. ## I made an error in problem

The second problem was actually supposed to be
f(x) - x ln square rt x

I will go back and rework it and see what happens

5. Hello, startingover!

f(x) .= .ln(2x)

Note that we have: .f(x) .= .ln(2) + ln(x)
. . . . . . . . . . . . . . . . . . . . . . . . . .
Then:. . . . . . . . . . f'(x) .= . . 0 . + .1/x

Therefore: .f'(x) .= .1/x