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January 30, 2015

January 30, 2015

Posted by **Coolio** on Sunday, June 30, 2013 at 6:04pm.

I know this is a combination of the chain rule and product rule. Please help!

- Calc I -
**Coolio**, Sunday, June 30, 2013 at 7:53pmNothing?

- Calc I -
**Steve**, Monday, July 1, 2013 at 12:28amYou know, if no one responds right away, a better follow-up would be to show whatever ideas you have had in the meantime, rather than just tapping your foot and saying "nothing?".

Your first thought, on seeing a complicated formula should have been the chain rule. what functions u and v could you use?

y = √u where

u = sin(e^(x^3)*cos(x))

u = sin v

where v = e^(x^3)*cos(x)

Now you can work with things:

y' = 1/2√u u'

= 1/2√u cos v v'

Now use the product rule:

= 1/2√u cos v e^(x^3) (3x^2 cosx - x^3 sinx)

Now just substitute back in for u and v.

- Calc I -
**Coolio**, Monday, July 1, 2013 at 1:07amThanks for your response. I was hardly taping my foot. I have been working on this problem for almost 6 days and by the time I posted this, I was over it. So I apologize if it seemed like I was impatient.

- Calc I/STEVE!!! -
**Coolio**, Monday, July 1, 2013 at 1:24amYou are the man!!!! I had tried this problem freaking 74 times. I had too many parenthesis, I forgot to add (e^(x^3)), and finally at the end of the derivative I did NOT include X^3 in front of sin(x).

Thank youuuuuuuu

- Calc I -
**Steve**, Monday, July 1, 2013 at 3:34amGood work. Many many times it is easier to make a few simple substitutions to avoid all the punctuation. Even the most complex problems contain simple parts.

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