# Derivatives of natural logarithmic functions

1. Sep 8, 2008

### fr33pl4gu3

f(x) = ln (12x-5/9x-2)

f'(x) = (4/3) (1/ln10)(9x-2/12x-5)

Is this correct??

2. Sep 8, 2008

### CompuChip

Instead of ln(10) you should put ln(base of the logarithm), in this case; ln(e) = 1. When deriving log(x) (base 10), you can rewrite to ln(x)/ln(10). ln(10) is just a constant, so the derivative of this is 1/ln(10) * 1/x.

Also, I think you didn't do the chain rule quite right.

3. Sep 8, 2008

### fr33pl4gu3

Then, is this
f(x) = ln (12x-5)
g(x) = ln (9x-2)
f'(x) = 12/12x-5
g'(x) = 9/9x-2

Correct??

4. Sep 8, 2008

### fr33pl4gu3

Is this correct:

(12/12(ln9x-2)-5)-(9/9x-2)

Last edited: Sep 8, 2008
5. Sep 8, 2008

### Defennder

This should not have ln inside. What is d/dx ln(12x-5) ? You've got it from above, just put it into this one.

6. Sep 9, 2008

### CompuChip

I have the feeling you are guessing more than working systematically. Actually there are two ways to solve this. One is by using the chain rule:
set y = (12x - 5) / (9x - 2). Then the derivative of ln(y) is 1/y . dy/dx.

The other way is to first use ln(a/b) = ln(a) - ln(b) and then use the sum rule to differentiate (you still need the chain rule!).