Chain Rule?

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  1. Sora's Avatar
    1 20-06-2012  18:02
    • Exalted Member
    • Location: Northampton
    • Posts: 251
    Chain Rule?
    Me again.

    Basically, I was taught the chain rule today along with the quotient and product rules and I could do all the question involving them except this one:

    Differentiate the following function:
    y = ln\sqrt{x^2-1}

    Thanks,
    Sora
  2. TenOfThem's Avatar
    2 20-06-2012  18:05
    • TSR Royalty
    Re: Chain Rule?
    Can you differentiate ln

    This is just the chain rule but with 3 functions
  3. EckoGecko's Avatar
    3 20-06-2012  18:13
    • Banned
    • Posts: 215
    Re: Chain Rule?
    I used chain rule twice, but not sure that's the best way to do it...
  4. Sora's Avatar
    4 20-06-2012  18:18
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    • Location: Northampton
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    Re: Chain Rule?
    3 functions?

    So far I have:

    y = ln(x^2-1)^{\frac{1}{2}} U = (x^2-1)^{\frac{1}{2}} y = lnU du/dx = 0.5(x^2-1)^{\frac{-1}{2}} dy/du = 1/x

    Is this right?
  5. samir12's Avatar
    5 20-06-2012  18:20
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    • Posts: 880
    Re: Chain Rule?
    (Original post by Sora)
    3 functions?

    So far I have:

    y = ln(x^2-1)^{\frac{1}{2}} U = (x^2-1)^{\frac{1}{2}} y = lnU du/dx = 0.5(x^2-1)^{\frac{-1}{2}} dy/du = 1/x

    Is this right?
    For du/dx, you forgot to multiply by 2x as (x^2 -1) differentiated is 2x and your dy/du should just be 1/U, not 1/x.
    Last edited by samir12; 20-06-2012 at 18:22.
  6. Sora's Avatar
    6 20-06-2012  18:21
    • Exalted Member
    • Location: Northampton
    • Posts: 251
    Re: Chain Rule?
    Is my "U" wrong? I think it's meant to be just (x^2 -1)
  7. james22's Avatar
    7 20-06-2012  18:23
    • Peer Of The TSR Realm
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    Re: Chain Rule?
    Your U is right, it's your dU/dx that is wrong. Apply the chain rule again to find du/dx.
  8. samir12's Avatar
    8 20-06-2012  18:26
    • Benevolent Member
    • Posts: 880
    Re: Chain Rule?
    (Original post by Sora)
    Is my "U" wrong? I think it's meant to be just (x^2 -1)
    It doesnt matter as long as you differentiate correctly, this question is more complex as you have 3 functions, within each other, the ln the (..)^1/2 and the x^2-1 so you have to use the chain rule when finding du/dx.
  9. Ashmanyeah's Avatar
    9 20-06-2012  18:29
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    • Posts: 35
    Re: Chain Rule?
    Isn't it just the differential of the equation in the brackets of ln over the original function of ln?
  10. james.h's Avatar
    10 20-06-2012  18:30
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    • Posts: 858
    Re: Chain Rule?
    Two-step chain rule (two substitutions)...

    Let u=x^2 - 1 and t=\sqrt{u}, then y=\ln(t).

    So (chain rule): \displaystyle \frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}t} \frac{\text{d}t}{\text{d}u} \frac{\text{d}u}{\text{d}x}
  11. TenOfThem's Avatar
    11 20-06-2012  18:34
    • TSR Royalty
    Re: Chain Rule?
    u = x^2-1

    v = u^{\frac{1}{2}}

    w = ln v


    \dfrac{dw}{dx} = \dfrac{dw}{dv} \times \dfrac{dv}{du} \times \dfrac{du}{dx}
  12. Sora's Avatar
    12 20-06-2012  19:16
    • Exalted Member
    • Location: Northampton
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    Re: Chain Rule?
    Thanks for all your help. I think I sort of combined the methods shown and figured it out.

    I did:
    u = (x^2-1)^{\frac{1}{2}} y = lnu

    then to find \dfrac{du}{dx} I used the chain rule again to get v=x^2-1 u=(v)^{\frac{1}{2}}
    and I used \dfrac{du}{dx}=\dfrac{dv}{dx} \dfrac{du}{dv} and then did:

    \dfrac{dy}{dx} = \dfrac{du}{dx}\dfrac{dy}{du}

    And got the right answer.

    Sorry if it is hard to follow xD
  13. TenOfThem's Avatar
    13 20-06-2012  19:29
    • TSR Royalty
    Re: Chain Rule?
    (Original post by Sora)
    Thanks for all your help. I think I sort of combined the methods shown and figured it out.

    I did:
    u = (x^2-1)^{\frac{1}{2}} y = lnu

    then to find \dfrac{du}{dx} I used the chain rule again to get v=x^2-1 u=(v)^{\frac{1}{2}}
    and I used \dfrac{du}{dx}=\dfrac{dv}{dx} \dfrac{du}{dv} and then did:

    \dfrac{dy}{dx} = \dfrac{du}{dx}\dfrac{dy}{du}

    And got the right answer.

    Sorry if it is hard to follow xD

    That is fine
  14. Sora's Avatar
    14 20-06-2012  19:40
    • Exalted Member
    • Location: Northampton
    • Posts: 251
    Re: Chain Rule?
    Thank you
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