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# Differentiate xsin(1/x)

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Reputation:
How do I differentiate xsin(1/x)
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Product rule. Chain rule. (1/x = x^(-1).)
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Reputation:
^^

Btw, I didn't think you could rewrite 1/x as x^-1 as an angle.
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Reputation:
So I got f'(x)=sin(1/x)-(1/x)cos(1/x)

Is this right at all?
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Ciaororwect,
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Reputation:
(Original post by giran)
^^

Btw, I didn't think you could rewrite 1/x as x^-1 as an angle.
Brrrrap learnin from TSR
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Reputation:
(Original post by giran)
^^

Btw, I didn't think you could rewrite 1/x as x^-1 as an angle.
It is a function...
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Reputation:
(Original post by s_abbott)
So I got f'(x)=sin(1/x)-(1/x)cos(1/x)

Is this right at all?
Spoiler:
Show

If y = x sin 1/x

u = x, v = sin 1/x
du/dx = 1, dv/dx = -1/(x^2) cos 1/x

u (dv/dx) + v (du/dx) = -1/x cos 1/x + sin 1/x

What I got...

EDIT: And I got it wrong
Changing it now.
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Reputation:
Innit man. Safe
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Reputation:
differentiate X, leave sin(1/x) alone. Multiply what you get. Thats 1 *sin(1/x).

then differentiate sin(x^-1) using the chain rule, and leave the X alone. Multiply what you get. So thats X * cos(x^-1) * -X^-2.

Then add the two results of the multiplication. so the end answer is:

1*sin(1/x) + X*cos(x^-1)*-X^-2

Clean it up and your laughing.
Elektrolite
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(Original post by Mathematician!)
Spoiler:
Show

If y = x sin 1/x

u = x, v = sin 1/x
du/dx = 1, dv/dx = 1/x cos 1/x

u (dv/dx) + v (du/dx) = cos 1/x + sin 1/x

What I got...
You need to account for the derivative of (1/x) for dv/dx.
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Reputation:
(Original post by DeanK2)
It is a function...
I know, it's just that I've hardly used that rule in exam questions. they give you nice numbers with a constant

and it'll become -1/x^2 * cos (1/x) when deriving sin(1/x)
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Reputation:
(Original post by Glutamic Acid)
You need to account for the derivative of (1/x) for dv/dx.
Yup, I was correcting it lol. Give me a chance
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Reputation:
(Original post by Mathematician!)
Spoiler:
Show

If y = x sin 1/x

u = x, v = sin 1/x
du/dx = 1, dv/dx = 1/x cos 1/x

u (dv/dx) + v (du/dx) = cos 1/x + sin 1/x

What I got...

EDIT: And I got it wrong
Changing it now.
you slipped up the dv/dx dude. remember, chain rule. differentiate the whole thing, (cos(1/x)) then times it by the differential of whats in the bracket (-X^-2).
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Reputation:
Well, I would have got that but I got:
u = x
du/dx = 1

v = sin(1/x)
dv/dx = (-1/x)cos(1/x)

Because 1/x = x^-1 so differentiating that gives -x^-2

Or am I going wrong??
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Reputation:
if v = sin(x^-1)

dv/dx = (-x^-2)cos(x^-1)
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Reputation:
(Original post by Glutamic Acid)
You need to account for the derivative of (1/x) for dv/dx.
OK it's now corrected. However the incorrected version is in your quote forever so I can get humiliated by it... FOREVER! NOOOO! Lol.
Ah well, I guess it proves I am human.
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Reputation:
Yes, check my last post.

dv/dx = -1/(x^2)*cos(1/x)

..... I really should learn the latex syntaxing.
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Reputation:
Hahaha cheers guys!!! Take care. Think I'll use TSR more, people are bloody helpful around here
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Reputation:
(Original post by Elektrolite)
you slipped up the dv/dx dude. remember, chain rule. differentiate the whole thing, (cos(1/x)) then times it by the differential of whats in the bracked (-X^-2).
ARGH! Lol I corrected it. Look at my edited version.

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