I just want to check another answer to see I am on the right track. I do not think I done this one correctly.

http://i244.photobucket.com/albums/g...g?t=1237847751

Printable View

- Mar 23rd 2009, 03:37 PMrobascUsing the chain rule
I just want to check another answer to see I am on the right track. I do not think I done this one correctly.

http://i244.photobucket.com/albums/g...g?t=1237847751 - Mar 23rd 2009, 03:46 PMPinkk
Totally wrong I'm afraid (I have no idea what you did). This requires the product rule and the chain rule.

Let

**The product rule:**

Applying the chain rule here:

You will need to apply the chain rule to , which ends up being which is simply . Getting back to the whole derivative we will have:

- Mar 23rd 2009, 07:05 PMrobasc
So, how did you determine that you had to use the product rule?

My book states nothing about learning to use the product rule to solve this problem?

I am very confused? - Mar 23rd 2009, 07:24 PMPinkk
You have a function such that , where and . If you have not learned the product rule yet, then your only other option is to expand and multiply every term by , and then differentiate.