# Logarithmic Differentiation

1. Jul 21, 2012

### surferbarney0729

The original problem is y = (2-x)^(sqrt x). If anyone who is rather confident with this could double check this example it would really help me out. Thanks. I attached the link to reduce my typing confusion.

http://archives.math.utk.edu/visual.calculus/3/logdiff.1/a1.gif

2. Jul 21, 2012

### SammyS

Staff Emeritus
Here's the image:

What is it that looks off to you?

#### Attached Files:

• ###### a1.gif
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3. Jul 21, 2012

### surferbarney0729

I can not pinpoint anything. I am aware of the steps of the log diff for other examples on this weeks homework. Prof through it out to the class that this looks off. I cant see it. Should I look deeper?

4. Jul 21, 2012

### skiller

Looks perfectly fine to me.

Provided you understand the product rule, the chain rule, and the fact that ln(a^b) = b.ln(a), there isn't really anything mysterious here at all.

5. Jul 21, 2012

### surferbarney0729

ok thanks. Perhaps it was thrown out to check our grasp of your aforementioned rules. Thanks for the time.

6. Jul 22, 2012

### surferbarney0729

I threw this out last night, but it is still getting some pushback.

How does this look? Right or a little off? I placed the problem and solution link since I can not figure out how to import the image.

http://archives.math.utk.edu/visual....gdiff.1/a1.gif [Broken]

Last edited by a moderator: May 6, 2017
7. Jul 22, 2012

### TheFool

This is what I get when trying to look at the image:

8. Jul 22, 2012

### surferbarney0729

9. Jul 22, 2012

### surferbarney0729

The original problem is...

y = (2-x)^sqrt(x)

10. Jul 22, 2012

### TheFool

And you want the derivative?

11. Jul 22, 2012

### surferbarney0729

yes, using logarithmic differentiation solve the problem.

12. Jul 22, 2012

### TheFool

In general, when you have something like $$f(x)^{g(x)}$$ you can find the derivative like this, which should give you what you need to solve your specific problem where log is the natural logarithm:

$$y=f(x)^{g(x)}$$
$$\log{y}=g(x)\log{f(x)}$$
Implicitly differentiate both sides:
$$\frac{dy}{y}=g'(x)\log{f(x)}dx+\frac{g(x)f'(x)}{f(x)}dx$$
$$y'=y\left(g'(x)\log{f(x)}+\frac{g(x)f'(x)}{f(x)} \right)$$
$$y'=f(x)^{g(x)}\left(g'(x)\log{f(x)}+\frac{g(x)f'(x)}{f(x)}\right)$$

13. Jul 22, 2012

### DonAntonio

It looks fine.

DonAntonio

14. Jul 23, 2012

### skiller

Your efforts to spread this question to a larger audience have obviously been quashed, as it looks like it's been merged into your original thread!
What exactly do you mean by "some pushback"? Has your Prof said something? If so, what? Is he/she worried about the domain/range not being real etc?

Even the God Almighty that is Woolfy-Alfa agrees that this is the correct solution.

Last edited: Jul 23, 2012