Deriving the Derivative of ln(x^2+2)^(1/2)

[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.

 

[Tuneman]

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

 

[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

 

[Tuneman]

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

 

[swears]

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

 

[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.

 

[swears]

Like This?

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

 

[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...

 

[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}

 

[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.)