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1018

  • one year ago

find df/dx and df/dy :

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  1. 1018
    • one year ago
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    \[\sqrt{4x ^{2} + 3y ^{2}}\]

  2. welshfella
    • one year ago
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    first simplify this a bit = [ (4x^2 + 3y^2)^½]^3 = (4x^2 + 3y^2)^(3/2)

  3. welshfella
    • one year ago
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    now you use the chain rule

  4. welshfella
    • one year ago
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    oh!! or is this partial differentiation??

  5. 1018
    • one year ago
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    wait, sorry, the derivatives are to be solved separately. like dx first then dy haha sorry

  6. 1018
    • one year ago
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    cause i saw that you should treat y as a constant etc etc for dx and vice versa for dy...

  7. welshfella
    • one year ago
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    yes - oh - its been a long time i'm struggling to remember this stuff...

  8. 1018
    • one year ago
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    haha, im just studying it right now and already struggling. lol. so, am i right?

  9. welshfella
    • one year ago
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    yes - you have the general idea but i'm a bit hazy on it to be honest

  10. 1018
    • one year ago
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    im just wondering if y is to be constant, so finding dx would just be sqrt(4x^2) dx ?

  11. 1018
    • one year ago
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    the radical is confusing me i might get it wrong

  12. welshfella
    • one year ago
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    i'm sure some of the other guys will be more familiar with it than i am.

  13. welshfella
    • one year ago
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    well chage the radical to an exponent as i have done

  14. 1018
    • one year ago
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    1/2 right?

  15. 1018
    • one year ago
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    ok i think i got it from here. haha. thanks!

  16. welshfella
    • one year ago
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    square root = ^(½) yes

  17. welshfella
    • one year ago
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    yw

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