# Q. f(x)=ln (12x-5/9x-2)So by using the chain rule, i can

1. Sep 9, 2008

### fr33pl4gu3

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

So by using the chain rule, i can get:

(-4/3)((9x-2)2/(12x-5)2)

and by using the quotient rule, i can get the final answer, which is:

(2(-36x-8)(-36x-15)2-2(-36x-15)(-36x-8)2)
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(-36x-15)4

The answer that i got here is wrong, but i don't know, why?

2. Sep 9, 2008

### NoMoreExams

Re: Differentiation4

First of all do you mean: $$f(x) = ln \left( \frac{12x-5}{9x-2} \right)$$? In which case yes you would use chain rule and quotient rule. Chain rule applies to cases like f(g(x)), what is your f? what is your g? let's start there.

3. Sep 9, 2008

### fr33pl4gu3

Re: Differentiation4

yes, and f is ln (12x-5/9x-2), where g is (12x-5/9x-2), correct??

4. Sep 9, 2008

### NoMoreExams

Re: Differentiation4

Perhaps I used too many f's, let's say we have

$$f(x) = g(h(x)) = ln \left( \frac{12x-5}{9x-2} \right)$$

What is g? what is h?

5. Sep 10, 2008

### fr33pl4gu3

Re: Differentiation4

Thanks, but i solve the problem, the f(x) will be the nominator of the log and the g(x) will be the denominator of the log.

6. Sep 10, 2008

### NoMoreExams

Re: Differentiation4

No, I asked about the chain rule, not the quotient rule. Actually f is the whole function, h(x) would be the argument of ln( ) and g would be ln itself. In general you should know this rule:

If

$$f(x) = ln( g(x) )$$

Then

$$f'(x) = \frac{g'(x)}{g(x)}$$

So in this case the derivative of

$$ln \left( \frac{12x-5}{9x-2} \right)$$

is

$$\frac{ \left(\frac{12x-5}{9x-2} \right)'} {\left( \frac{12x-5}{9x-2} \right) }$$

Where ' denotes differentiation.

7. Sep 11, 2008

### HallsofIvy

Staff Emeritus
Re: Differentiation4

Since this doesn't really have anything to do with differential equations, I am moving it to Calculus and Analysis.