Differentiating ln(2^x+3^x)

  • Thread starter jokerzz
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  • #1
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I think I know how to do this but i'm completely blank at the moment. Do I differentiate 'ln' and then differentiate 2^x and 3^x in the denominator?
 

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  • #2
Mentallic
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By the chain rule, we have if [tex]y=ln(f(x))[/tex] then [tex]\frac{dy}{dx}=\frac{1}{f(x)}f'(x)[/tex]
 
  • #3
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ok so now I hav a bigger problem. The original question was that I needed to find the limit as n approaches infinity, ln(2^n+3^n)/n approaches...?? I know I need to use the lopital rule but im not getting anywhere with it
 
  • #4
dextercioby
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Hint

[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

And now use the logarithm's properties.
 
  • #5
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ok so that worked but when I take 2^n common i get the answer "ln(2)+ln(1.5)" but wen i take 3^n common, i get "ln(3)+ln(2/3)" and these are different answers
 
  • #6
dextercioby
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You';re making a mistake somewhere. Use the fact that:

[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]
 
Last edited:
  • #7
VietDao29
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Hint

[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

Err.. Sorry, but I couldn't get your point... What are you trying to do?

And now use the logarithm's properties.
ok so that worked but when I take 2^n common i get the answer "ln(2)+ln(1.5)" but wen i take 3^n common, i get "ln(3)+ln(2/3)" and these are different answers
When dealing with taking the derivatives of composite functions f(u(x)), one should immediately think of Chain Rule. The Chain Rule states that:

[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex]

I'll give you an example so that you can understand how to apply the rule.
Example:
Differentiate: f(x) = sin(x2).
--------------------
Here, we notice that, if we define another function u(x) = x2, our function will become: f(x) = sin(x2) = sin(u(x)). Which is a composite function.

So, by using the Chain Rule, we have:

[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx} = \frac{d(\sin u)}{du} \times \frac{d(x ^ 2)}{dx} = \cos(u) \times 2x = 2x\cos(x ^ 2)[/tex].

Let's see if you can tackle your problem. It's pretty much the same as the example above. :)
 
  • #8
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You';re making a mistake somewhere.

[tex] \lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0 [/tex]

It is?? How come? I took it as "(2/3)^n"
 
  • #9
dextercioby
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It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.
 
  • #10
VietDao29
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Hint

[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(3^n\left(1+\frac{2^n}{3^n}\right)\right)}{n} [/tex]

And now use the logarithm's properties.
May I ask, what does taking this limit have anything to do with differentiating y = ln(3x + 2x)? :confused:
 
  • #11
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May I ask, what does taking this limit have anything to do with differentiating y = ln(3x + 2x)? :confused:
The question I originally asked was just part of the whole question. Read my post above
 
  • #12
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It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.
Thanks dude for your help. I understand now
 
  • #13
dextercioby
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He was asked to find the value of the limit

[tex] \lim_{n\rightarrow\infty} \frac{\ln\left(2^n + 3^n \right)}{n} [/tex].

First he though of setting the fraction as a function and using l'Ho^pital's rule. I only tried to show him that he needn't know any differentiation to solve his problem.
 
  • #14
HallsofIvy
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May I ask, what does taking this limit have anything to do with differentiating y = ln(3x + 2x)? :confused:
The OP wanted to differentiate that in order to use L'Hopital's rule to do that limit. As usual, there was a far easier way to find the limit than to use L'Hopital's rule. Another case of not asking the question you really want answered!
 
  • #15
VietDao29
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Whoops... My bad. I dun really know why, but I just completely missed post #3... Sorry about that. :(
 

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