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Partial differentiation

  1. Jan 8, 2007 #1
    Iv forgotten the basics of this. How do we go about differentiating 3x^2 y^2 w.r.t y? I know the answer is 3x^2 y^2 but could someone explain this for me please?
  2. jcsd
  3. Jan 8, 2007 #2


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    You want to perform [tex]\frac{\partial}{\partial y}(3x^2y^2)[/tex].

    You would simply treat the function of x as a constant and differentiate y normally, thus yielding [itex]6x^2y[/itex].

    However, this is not what your answer is, so I wonder whether it is the correct question?
  4. Jan 8, 2007 #3
    ooops im getting myself all confused now. I know how to do partial differentiation after all. I was reading the wrong solution as the question was z = x^2 y^2 and ofcourse dz/dy would = 3x^2 y^2 !!
  5. Jan 8, 2007 #4


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    Erm.. no it wouldn't.

    Unless you mean z=x2y3, in which case, dz/dy would be the answer you quoted.
  6. Jan 8, 2007 #5
    i need some sleep!! Yeah i meant y^3 lol !
  7. Jan 8, 2007 #6

    Gib Z

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    Its a big warning flag when the derivative is equal to the original function and doesn't involve e in it....
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