C1 Differentiation

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  1. krisshP's Avatar
    1 29-07-2012  15:24
    • Peer Of The TSR Realm
    • Posts: 1,678
    C1 Differentiation
    I'm stuck on the question which is below:

    If y=\sqrt{x}, show that 2x\frac{dy}{dx}=y

    Below is my working out:

    y=x^\frac{1}{2}

    This differentiates to provide:
    \frac{dy}{dx}=(\frac{1}{2}x)^-\frac{1}{2}

    I multiply the above by 2x to get:
    (x^3)^-\frac{1}{2}

    x^-\frac{3}{2}

    What have I done wrong?????:confused::confused::confused:

    Thanks for any help provided.
  2. boromir9111's Avatar
    2 29-07-2012  15:28
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    • Location: Here and There
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    Re: C1 Differentiation
    (Original post by krisshP)
    I'm stuck on the question which is below:

    If y=\sqrt{x}, show that 2x\frac{dy}{dx}=y

    Below is my working out:

    y=x^\frac{1}{2}

    This differentiates to provide:
    \frac{dy}{dx}=(\frac{1}{2}x)^-\frac{1}{2}

    I multiply the above by 2x to get:
    (x^3)^-\frac{1}{2}

    x^-\frac{3}{2}

    What have I done wrong?????:confused::confused::confused:

    Thanks for any help provided.
    dy/dx = 1/[2(x)^(1/2)]

    2x*dy/dx = 2x/2x^(1/2)

    laws of indices......x^1 - x^(1/2)

    2's cancel out and you know y = x^1/2
  3. krisshP's Avatar
    3 29-07-2012  15:46
    • Peer Of The TSR Realm
    • Posts: 1,678
    Re: C1 Differentiation
    (Original post by boromir9111)
    dy/dx = 1/[2(x)^(1/2)]

    2x*dy/dx = 2x/2x^(1/2)

    laws of indices......x^1 - x^(1/2)

    2's cancel out and you know y = x^1/2
    How do you get the following:

    dy/dx = 1/[2(x)^(1/2)]

    :confused::confused:

    I do not understand.
    btw can you please use latex.

    Thanks
  4. boromir9111's Avatar
    4 29-07-2012  15:50
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    • Location: Here and There
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    Re: C1 Differentiation
    (Original post by krisshP)
    How do you get the following:

    dy/dx = 1/[2(x)^(1/2)]

    :confused::confused:

    I do not understand.
    btw can you please use latex.

    Thanks
    No worries

    \displaystyle\frac{dy}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-\frac{1}{2}}

    Your method of calculating dy/dx is correct but it's better to leave it in fraction form as it's more neater!
    Last edited by boromir9111; 29-07-2012 at 15:51.
  5. Lord of the Flies's Avatar
    5 29-07-2012  15:51
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: C1 Differentiation
    (Original post by krisshP)
    I'm stuck on the question which is below:

    If y=\sqrt{x}, show that 2x\frac{dy}{dx}=y

    Below is my working out:

    y=x^\frac{1}{2}

    This differentiates to provide:
    \frac{dy}{dx}=(\frac{1}{2}x)^-\frac{1}{2}

    I multiply the above by 2x to get:
    (x^3)^-\frac{1}{2}

    x^-\frac{3}{2}

    What have I done wrong?????:confused::confused::confused:

    Thanks for any help provided.
    Hint:

    \dfrac{dy}{dx}=\dfrac{1}{2} x^{-\frac{1}{2}}\iff \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{ x}}\iff \dfrac{dy}{dx}2\sqrt{x}=1

    Edit: woops I'm a bit late!
    Last edited by Lord of the Flies; 29-07-2012 at 15:52.
  6. krisshP's Avatar
    6 29-07-2012  16:00
    • Peer Of The TSR Realm
    • Posts: 1,678
    Re: C1 Differentiation
    2x(\frac{1}{2}x)^{-\frac{1}{2}}

    (\frac{2x^2}{2})^{-\frac{1}{2}}

    (x^2)^{-\frac{1}{2}}

    x^{-1}

    What have I now done wrong???????:confused:

    btw thanks for the tip on latex.
  7. Lord of the Flies's Avatar
    7 29-07-2012  16:05
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: C1 Differentiation
    (Original post by krisshP)
    2x(\frac{1}{2}x)^{-\frac{1}{2}}

    (\frac{2x^2}{2})^{-\frac{1}{2}}

    (x^2)^{-\frac{1}{2}}

    x^{-1}

    What have I now done wrong???????:confused:

    btw thanks for the tip on latex.
    Your first mistake is (\dfrac{1}{2}x)^{-\frac{1}{2}}

    It should be \dfrac{1}{2}x^{-\frac{1}{2}}
  8. boromir9111's Avatar
    8 29-07-2012  16:06
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    • Location: Here and There
    • Posts: 10,801
    Re: C1 Differentiation
    (Original post by krisshP)
    2x(\frac{1}{2}x)^{-\frac{1}{2}}

    (\frac{2x^2}{2})^{-\frac{1}{2}}

    (x^2)^{-\frac{1}{2}}

    x^{-1}

    What have I now done wrong???????:confused:

    btw thanks for the tip on latex.
    If you insist on doing it this way then you should be careful when applying laws of indices here

    x^1*x^(-1/2) does not give x^2
  9. TenOfThem's Avatar
    9 29-07-2012  16:09
    • TSR Royalty
    Re: C1 Differentiation
    (Original post by krisshP)
    2x(\frac{1}{2}x)^{-\frac{1}{2}}

    (\frac{2x^2}{2})^{-\frac{1}{2}}

    (x^2)^{-\frac{1}{2}}

    x^{-1}

    What have I now done wrong???????:confused:

    btw thanks for the tip on latex.

    You are not using the rules of indices correctly

    \frac{dy}{dy} = \frac{1}{2}x^{-\frac{1}{2}}

    is your first error
  10. TenOfThem's Avatar
    10 29-07-2012  16:12
    • TSR Royalty
    Re: C1 Differentiation
    Then 2x \times \frac{1}{2}x^{-\frac{1}{2}} = x \times x^{-\frac{1}{2}}

    Then 1-\frac{1}{2} = \frac{1}{2}
    Post rating:
    1
  11. krisshP's Avatar
    11 29-07-2012  16:16
    • Peer Of The TSR Realm
    • Posts: 1,678
    Re: C1 Differentiation
    (Original post by TenOfThem)
    Then 2x \times \frac{1}{2}x^{-\frac{1}{2}} = x \times x^{-\frac{1}{2}}

    Then 1-\frac{1}{2} = \frac{1}{2}
    Thanks a lot for your support.

    I understand now.
    Post rating:
    1
  12. TenOfThem's Avatar
    12 29-07-2012  16:17
    • TSR Royalty
    Re: C1 Differentiation
    (Original post by krisshP)
    Thanks a lot for your support.

    I understand now.
    ok good
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