# Another differentiation question.

1. Apr 28, 2005

### monet A

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!

2. Apr 28, 2005

### OlderDan

Yes you can do it. Don't forget your left side is lny and not y

3. Apr 28, 2005

### cepheid

Staff Emeritus
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. Apr 28, 2005

### monet A

Thanks again both of you, and yeah cepheid my foundation stuff is a bit scratchy thanks for the tip on that log rule.

5. Apr 28, 2005

### monet A

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?

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: Apr 28, 2005
6. Apr 28, 2005

### Data

$$\log (a+b) \neq \log a + \log b$$

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

7. Apr 28, 2005

### monet A

Ahhh thankyou, I will be reviewing my Log rules now. :tongue2:

8. Apr 28, 2005

### monet A

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. Apr 28, 2005

### Data

looks just fine to me.