Another differentiation question.

I'm not sure what you mean by "necessary," but if it works, it works. In summary, the conversation discusses logarithmic differentiation and the application of the product rule to g(x) * ln(f(x)). The conversation also touches on log laws and the proper way to split up logarithmic expressions.
  • #1
monet A
67
0
Hi again,
I am in the middle of a logarithmic differentiation and I have

y = f(x)^g(x)

ln y = g(x) * ln(f(x))

Can I just apply the product rule to g(x) * ln(f(x)) or am I going to mess with something, I am drawing a blank on log rules that I might be violating.
Thanks!
 
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  • #2
monet A said:
Hi again,
I am in the middle of a logarithmic differentiation and I have

y = f(x)^g(x)

ln y = g(x) * ln(f(x))

Can I just apply the product rule to g(x) * ln(f(x)) or am I going to mess with something, I am drawing a blank on log rules that I might be violating.
Thanks!

Yes you can do it. Don't forget your left side is lny and not y
 
  • #3
It looks fine. If you want to differentiate both sides, doing the RHS using the product rule...go ahead. After all, you have two functions multiplied.

But you didn't seem to be doubting the product rule so much as the log laws that led to the second equation. Well, if y = f^g, and you want to find the natural log of y, you are finding the exponent needed to express y as a power of e. That would be the exponent needed to express f^g as a power of e. But when you raise something to a power (g in this case) you just multiply exponents. So that would be g times the exponent needed to express f as a power of e. That's a good way of remembering that log rule.
 
  • #4
Thanks again both of you, and yeah cepheid my foundation stuff is a bit scratchy thanks for the tip on that log rule.
 
  • #5
While I am on the subject of Logs I have another question.

When I'm taking logs of both sides of y = [(x^0.5 +x)^)0.5 + x]^0.5
in other words y = sqrt[x+sqrt{x+sqrt(x)}]

Do I do the RHS like this: 1/2 [ ln(x^0.5 + x)^0.5 + lnx ] or can I not take the log of that last x separately?

:blushing: Yes I really am very rusty on the rule I am applying.

By the way, Should I move my questions to the calculus forum?
 
Last edited:
  • #6
[tex]\log (a+b) \neq \log a + \log b[/tex]

so no, you can't split it up that way.
 
  • #7
Data said:
[tex]\log (a+b) \neq \log a + \log b[/tex]

so no, you can't split it up that way.


Ahhh thankyou, I will be reviewing my Log rules now. :-p
 
  • #8
I broke it up like this :

let sqrt(sqrt(x)) = g(x) and +x = f(x)

then ln y = 1/2 [ln {g(x) + f(x)}]

and since d/dx[f(x)+/-g(x)] = df(x)/dx +/- dg(x)/dx

then 1/y*dy/dx = 1/2 [{df(x)/dx + dg(x)/dx)}/{f(x) + g(x)}]

It worked but was that necessary or did I just sledgehammer a peanut?
 
  • #9
looks just fine to me.
 

Related to Another differentiation question.

1. What is differentiation?

Differentiation is a mathematical process used to find the rate of change of a function with respect to one of its variables. It is commonly used in calculus to solve problems involving slopes, velocities, and rates of change.

2. How is differentiation different from integration?

Differentiation and integration are inverse processes. While differentiation calculates the rate of change of a function, integration calculates the total value of a function over a given interval.

3. What is the chain rule in differentiation?

The chain rule is a formula used to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

4. Can differentiation be used in real-world applications?

Yes, differentiation has many practical applications in fields such as physics, engineering, economics, and statistics. It is used to calculate rates of change, optimize functions, and model real-world phenomena.

5. Are there any limitations to differentiation?

One limitation of differentiation is that it cannot be used to find the slope of a function at a point where the function is not continuous. Additionally, some functions may not have a well-defined derivative at certain points, making it impossible to use differentiation in those cases.

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