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

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  • #1
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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?
 

Answers and Replies

  • #2
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First of all do you mean: [tex] f(x) = ln \left( \frac{12x-5}{9x-2} \right) [/tex]? 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
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yes, and f is ln (12x-5/9x-2), where g is (12x-5/9x-2), correct??
 
  • #4
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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

[tex] f(x) = g(h(x)) = ln \left( \frac{12x-5}{9x-2} \right) [/tex]

What is g? what is h?
 
  • #5
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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
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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

[tex] f(x) = ln( g(x) ) [/tex]

Then

[tex] f'(x) = \frac{g'(x)}{g(x)}[/tex]

So in this case the derivative of

[tex] ln \left( \frac{12x-5}{9x-2} \right) [/tex]

is

[tex] \frac{ \left(\frac{12x-5}{9x-2} \right)'} {\left( \frac{12x-5}{9x-2} \right) } [/tex]

Where ' denotes differentiation.
 
  • #7
HallsofIvy
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Since this doesn't really have anything to do with differential equations, I am moving it to Calculus and Analysis.
 

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