How do you differentiate ln(2^x+3^x)?

[jokerzz]

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?

 

[Mentallic]

By the chain rule, we have if y=ln(f(x)) then \frac{dy}{dx}=\frac{1}{f(x)}f'(x)

 

[jokerzz]

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 I am not getting anywhere with it

 

[dextercioby]

Hint

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

And now use the logarithm's properties.

 

[jokerzz]

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

 

[dextercioby]

You';re making a mistake somewhere. Use the fact that:

\lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0

 

[VietDao29]

Hint

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

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:

\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}

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:

\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).

Let's see if you can tackle your problem. It's pretty much the same as the example above. :)

 

[jokerzz]

You';re making a mistake somewhere.

\lim_{n\rightarrow\infty} \frac{2^n}{3^n} = 0

It is?? How come? I took it as "(2/3)^n"

 

[dextercioby]

It's the same thing. 2/3 is under unity. Multiplying it an infinite nr of times gives 0.

 

[VietDao29]

Hint

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

And now use the logarithm's properties.

May I ask, what does taking this limit have anything to do with differentiating y = ln(3^x + 2^x)?

 

[jokerzz]

May I ask, what does taking this limit have anything to do with differentiating y = ln(3^x + 2^x)?

The question I originally asked was just part of the whole question. Read my post above

 

[jokerzz]

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

 

[dextercioby]

He was asked to find the value of the limit

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

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.

 

[HallsofIvy]

May I ask, what does taking this limit have anything to do with differentiating y = ln(3^x + 2^x)?

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!

 

[VietDao29]

Whoops... My bad. I dun really know why, but I just completely missed post #3... Sorry about that. :(