# Derivaties of trigonometric functions and natural logs.

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• Nov 21st 2010, 08:25 PM
FullBox
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Derivaties 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, 08:51 PM
TheCoffeeMachine
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, 09:19 PM
harish21
Quote:

Originally Posted by FullBox
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.

The differentiation is correct for the second one. You can cancel off the 13 in the numerator and the denominator and make it $\dfrac{1}{x \ln(13)}$
• Nov 22nd 2010, 05:38 AM
FullBox
Quote:

Originally Posted by TheCoffeeMachine
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.

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.

Quote:

Originally Posted by harish21
The differentiation is correct for the second one. You can cancel off the 13 in the numerator and the denominator and make it $\dfrac{1}{x \ln(13)}$

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, 05:44 AM
HallsofIvy
Quote:

Originally Posted by FullBox
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.

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 $5x cos(x^2)$. The $x^2$ 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 $cos(x^2)$ for which you want to use the product rule and then a composition: $cos(x^2)= cos(y)$ with $y= x^2$.

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, 04:57 PM
FullBox
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, 05:04 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
So what would I do after applying the product rule?

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.
You're one of the first few people to say that, and for that reason kudos to you! (Clapping)
• Nov 22nd 2010, 05:20 PM
FullBox
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, 05:58 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
Would this be the correct first take step to take?

You sort of misunderstand the chain rule. It says that $[f(g(x))]' = f'(g(x))g'(x)$, it doesn't say
$[f(g(x))]' = f'(g(x))+g'(x)$. Let me give you an example: $[\tan(x^3)]' = \left(\sec^2(x^3)\right)(3x^2)$.
• Nov 22nd 2010, 06:06 PM
FullBox
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, 06:14 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
Wow total flop on my part, my mind was still partially thinking of the product rule.

Close but not quite yet. Hint: $f'(x^2) \ne \frac{d}{dx}\left(\cos(x^2)\right) \ne -2\sin{x}$.

Look at my example again: $f'(x^3) = \tan'(x^3) = \sec^2(x^3)$.
• Nov 22nd 2010, 06:25 PM
FullBox
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, 06:29 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
So I don't touch whats inside the parenthesis, just the outer part of the function?

Yep! That's it. You're missing the square from the x² in the argument of sine in the end, but I'll assume
it's just a typo. Now go back to your very original product $5x \times \cos(x^2)$ and apply the product rule.
• Nov 22nd 2010, 06:42 PM
FullBox
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, 06:50 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
Yeah that was definitely a typo, thanks for pointing it out. I corrected it in this post.

Nice. That's the first derivative! (Yes)

Now, of course you need to find the second derivative. Differentiate what you have got using the chain &
product rules again, of course, and you are done. I'm sure you can do it easily now that you understand.
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