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    Please can any maths people try to differentiate the following equation for me: y = -1/(x^4-8), where ^ means 'to the power'. Thanks a lot. :confused:
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    (Original post by play_the_world)
    Please can any maths people try to differentiate the following equation for me: y = -1/(x^4-8), where ^ means 'to the power'. Thanks a lot. :confused:
    By the quotient rule
    y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

    u = -1 => du/dx = 0
    v = [x4 - 8] => dv/dx = 4x³

    dy/dx
    = [(x4 - 8) x 0 - (-1 x 4x³)]/(x4 - 8)²
    = (0 + 4x3)/(x4 - 8)²
    = 4x³/(x4 - 8)²

    By the chain rule
    Let u = x4 - 8 => du/dx = 4x³
    => y = -1/u => dy/du = 1/u²

    dy/dx
    = dy/du x du/dx
    = 1/u² x 4x³
    = 4x³/u²
    = 4x³/(x4 - 8)²
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    (Original post by Widowmaker)
    By the quotient rule
    y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

    u = -1 => du/dx = 0
    v = [x4 - 8] => dv/dx = 4x³

    dy/dx
    = [(x4 - 8) x 0 - (-1 x 4x³)]/(x4 - 8)²
    = (0 + 4x3)/(x4 - 8)²
    = 4x³/(x4 - 8)²

    By the chain rule
    Let u = x4 - 8 => du/dx = 4x³
    => y = -1/u => dy/du = 1/u²

    dy/dx
    = dy/du x du/dx
    = 1/u² x 4x³
    = 4x³/u²
    = 4x³/(x4 - 8)²
    Thankyou for confirming the answer. That's what i got too! And such a quick response x x x
 
 
 
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