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

    Btw, I didn't think you could rewrite 1/x as x^-1 as an angle.
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    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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    (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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    (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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    (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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    Innit man. Safe
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    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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    (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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    (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 :huff:
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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...

    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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    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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    if v = sin(x^-1)

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

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

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