# Chain Rule Help

1. Jun 18, 2006

### swears

$$y = ln[\sqrt(x^2+2)]$$

I changed it to $$y = ln[(x^2+2)^{1/2}]$$

I know the derivative of $$ln = \frac {1}{x}$$

I'm not sure how to proceed.

Last edited by a moderator: Jun 19, 2006
2. Jun 18, 2006

### Tuneman

set the argument of the natural log equal to u. Take derivative of u!

3. Jun 18, 2006

### swears

If I knew how to take the derivative I wouldn't have asked!

The way it looks to me I get:

$$\frac {1}{(x^2 + 2)^{1/2}} * \frac{1}{2}(x^2 + 2)^{-1/2}$$

I'm not sure this is right though

Last edited: Jun 18, 2006
4. Jun 18, 2006

### Tuneman

you have to usea certain method, perhaps a certain rule that you have been studying...

5. Jun 18, 2006

### swears

So i'm guessing the rule in the topic is wrong then?

6. Jun 18, 2006

### Tuneman

what I mean is you need to do the chain rule one more time. this is an example of doing multiple chain rules in 1 problem.

7. Jun 18, 2006

### swears

Like This?

$$\frac {1}{(x^2 + 2)^{1/2}} * \frac{-1}{4}(x^2 + 2)^{-1.5} (2x)$$

8. Jun 18, 2006

### EbolaPox

May I make a suggestion to ease the diifficulty? Here's ahint

$$ln(a^b) = bln(a)$$

That should deal with the square root. Then use a u substitution on the x^2 + 2 inside. That may simplify the problem a bit.

Edit: Oops, mistyped that logarithm...

9. Jun 18, 2006

### swears

So I can change it to this:

$$y = \frac {1}{2}ln(x^2+2)$$

Then Use the Product Rule Instead?

If so, I get:

$$y' = 1x \frac {1x}{x^2 +2}$$

Last edited: Jun 18, 2006
10. Jun 19, 2006

### HallsofIvy

Product rule? What product? If you mean the "2" times "ln...", since the derivative of a constant, normally you don't use the product rule in a situation like that: (2f(x))'= 2f'(x). But, you did get the correct answer!
(I might be inclined not to write the 1s.)