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#1 2014-04-15 03:16:07

Shelled
Guest

Derivatives.

Find the derivative of
d/dx [(sin x)^(1/x)]

I don't think this is right, but here's my answer. I used the quotient & chain rule to get

(x^2 sqrt(x^2+7))/((x^5-3x^3+1)*sqrt(1-2x^3)) + (x sqrt (x^5 +3x^3+1)/((3 sqrt (x^2+7))(x^5-3x^2+1))

#2 2014-04-15 03:56:59

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Hi;

Is this your problem:


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#3 2014-04-15 05:08:34

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Sorry, hopefully this is more clear (the question, with some of my working out)


i.imgur.com/VyT9EPC.jpg

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#4 2014-04-15 05:33:24

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,097

Re: Derivatives.

hi Shelled

Welcome to the forum.

Thanks for the image.  You seem to have three different problems muddled up there.  Let's try this one:

Treat this as:

So this needs the chain rule only.

Hope that helps.  smile  If that wasn't the problem, or you want more help, please post again.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#5 2014-04-15 10:00:47

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Whoops, there's a typo in the question; it's supposed to be

Could I still use the chain rule?

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#6 2014-04-15 10:03:35

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Hi;

That is what I asked you in post #2.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#7 2014-04-15 10:09:11

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Okay, sorry it's been a long day. I just realized that the working out I put up was for a different question too hmm

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#8 2014-04-15 10:16:01

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

If you want to use the chain rule what would you pick?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#9 2014-04-15 10:27:39

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Actually, not the chain rule.
Would I use the quotient rule?

Edit: wait. I'll stick with the chain rule smile I think I might know how to solve this. I'll post the answer in a bit.

Last edited by Shelled (2014-04-15 10:32:27)

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#10 2014-04-15 10:38:47

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Hi;

Okay, post when you have it.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#11 2014-04-15 10:58:14

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Okay, so not the complete answer, but am I going in the right direction?

and then apply the chain rule?

where 


   <--- not sure how to get the derivative of this. I think it's the product rule where...
                 
       and       

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#12 2014-04-15 11:13:49

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

So far you are correct so see if you can go a little farther.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#13 2014-04-15 11:46:01

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

So I used the product rule; then applied it all to the chain rule and got

i.imgur.com/McRIMWE.png

having trouble with simplifying it though

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#14 2014-04-15 12:13:15

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Hi Shelled;

That is correct! You could do a little simplifying though.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#15 2014-04-15 12:19:23

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Thankyou smile Got the right answer

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#16 2014-04-15 12:23:08

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Is that the answer they wanted?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#17 2014-04-15 12:34:28

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

I managed to simplify it to get the answer

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#18 2014-04-15 12:39:44

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,472

Re: Derivatives.

Okay, very good.  Welcome to the forum.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#19 2014-04-15 12:50:15

Shelled
Member
Registered: 2014-04-15
Posts: 26

Re: Derivatives.

Thanks smile

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