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

[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)
------------------------------------------------------------------
(-36x-15)⁴

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

 

[NoMoreExams]

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.

 

[fr33pl4gu3]

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

 

[NoMoreExams]

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?

 

[fr33pl4gu3]

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.

 

[NoMoreExams]

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]

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