Hello everyone. I am in need of help for some of my online practice problems. I typed them up on Microsoft Word as well as my work, and was hoping for further assistance.

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- Nov 21st 2010, 07:25 PMFullBoxDerivaties of trigonometric functions and natural logs.
Hello everyone. I am in need of help for some of my online practice problems. I typed them up on Microsoft Word as well as my work, and was hoping for further assistance.

http://www.mathhelpforum.com/math-he...isc/pencil.png - Nov 21st 2010, 07:51 PMTheCoffeeMachine
For the first one, it isn't the composition of three functions you have, it's the product of two functions, 5x

and cos(x²), one of which is the composition of two functions (the latter). Also you'll need to find the 2nd

derivative. For the second one, correct. Well done. Just cancel the 13's for simplification's sake, perhaps. - Nov 21st 2010, 08:19 PMharish21
- Nov 22nd 2010, 04:38 AMFullBox
I'm just wondering, how do you find out what the composition actually is? That is probably my biggest problem; figuring out how I should split up the problem.

Thank you. I'm always worried when I simplify because sometimes I don't replace the values with something equivalent and mess everything up. - Nov 22nd 2010, 04:44 AMHallsofIvy
Are you clear on exactly what a "composition" of two functions is? It means something of the form f(g(x)), not just any combination of functions like, say f(x)g(x).

Here, you had the function $\displaystyle 5x cos(x^2)$. The $\displaystyle x^2$ is**inside**the cos( ) so that is a composition. The "5x" and "cos(x)" are not**inside**any function of a single variable, they are multiplied together. That is why it is wrong to say you have a "composition of three functions". You have the product of two functions, 5x, and $\displaystyle cos(x^2)$ for which you want to use the product rule and**then**a composition: $\displaystyle cos(x^2)= cos(y)$ with $\displaystyle y= x^2$.

Thank you. I'm always worried when I simplify because sometimes I don't replace the values with something equivalent and mess everything up.[/QUOTE] - Nov 22nd 2010, 03:57 PMFullBox
So what would I do after applying the product rule? Please do not give me the answer. Rather nudge me in the right direction.

http://oi51.tinypic.com/2gxlpcm.jpg - Nov 22nd 2010, 04:04 PMTheCoffeeMachine
You applied the product rule to the composite function whereas you should have used the chain rule.

Quote:

Please do not give me the answer. Rather nudge me in the right direction.

- Nov 22nd 2010, 04:20 PMFullBox
I really need to understand this material, and if you just gave me the answer it'd be useless.

http://oi53.tinypic.com/35idirs.jpg

Would this be the correct first take step to take? - Nov 22nd 2010, 04:58 PMTheCoffeeMachine
- Nov 22nd 2010, 05:06 PMFullBox
Wow total flop on my part, my mind was still partially thinking of the product rule.

http://oi52.tinypic.com/2hhgpeg.jpg

But now, wouldn't I be in a similar situation as the original function? This new function derived from the chain rule is of very similar format to the original function. - Nov 22nd 2010, 05:14 PMTheCoffeeMachine
- Nov 22nd 2010, 05:25 PMFullBox
So I don't touch whats inside the parenthesis, just the outer part of the function?

http://oi55.tinypic.com/6toh0l.jpg - Nov 22nd 2010, 05:29 PMTheCoffeeMachine
- Nov 22nd 2010, 05:42 PMFullBox
Yeah that was definitely a typo, thanks for pointing it out. I corrected it in this post.

http://oi53.tinypic.com/t01eeu.jpg - Nov 22nd 2010, 05:50 PMTheCoffeeMachine