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# Differentiate xsin(1/x)

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Why bother with a post grad course - waste of time? 17-10-2016
1. How do I differentiate xsin(1/x)
2. Product rule. Chain rule. (1/x = x^(-1).)
3. ^^

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

Is this right at all?
5. Ciaororwect,
6. (Original post by giran)
^^

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

Btw, I didn't think you could rewrite 1/x as x^-1 as an angle.
It is a function...
8. (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.
9. Innit man. Safe
10. 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
11. (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.
12. (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)
13. (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
14. (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).
15. 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??
16. if v = sin(x^-1)

dv/dx = (-x^-2)cos(x^-1)
17. (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.
18. Yes, check my last post.

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

..... I really should learn the latex syntaxing.
19. Hahaha cheers guys!!! Take care. Think I'll use TSR more, people are bloody helpful around here
20. (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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