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Homework Help: Hiya! Need HELP?

  1. Apr 18, 2005 #1
    hay everyone....need some help on the question :cry:

    I need to find dz/dx in terms of x and y, if:

    Z = x^2y - 2xy^2 + y - 8 and y = 2x^2 +3/x!

    tried solving it a few times, but its a difficult one! any help and working out would be great! :confused:
  2. jcsd
  3. Apr 18, 2005 #2


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    We'd be better able to help if you show us what you've done.
  4. Apr 18, 2005 #3
    Ok dz/dx

    x = 2y - 4y +y

    y = 4x + 3

    dont think that is right!
  5. Apr 18, 2005 #4


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    You are given z as a function of x and y, and y as a function of x. So I suggest that you substitute all the y int he equation of z for their value in terms of x and then calculate dz/dx.
  6. Apr 18, 2005 #5


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    Use the chain rule. Since y is a function of x, dz/dx= 2xy+ x2y'- 2y2+ 4xy6y' + y'. since y = 2x2 +3/x, you can find y' as a funcion of x and substitute.
  7. Apr 21, 2005 #6
    i think i can help,dz/dx=x^2y[dy/dx(log[x^2])+2y/x]-2y^2-4xydy/dx+dy/dx
    where dy/dx=4x - {3/x!}[1/x+1/x-1+1/x-2+1/x-3.......]
  8. Apr 22, 2005 #7
    hallsofivy ur step was wrong since y is a function of x cant differentiate like that
    check ur solution one more time,ur step will be right if the question reads x^2 times y ,but if it is x raised to the power of 2y then mine is correct.also i hope the 6 in the function 4xy6y' was a typing mistake cos if it is not then ur answer is wrong.
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