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

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?

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.

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

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

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.

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.

HallsofIvy