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Partial derivative

  1. Sep 12, 2013 #1
    Hi everyone,


    I would like to find the partial derivative of:
    diff(z,x) ?

    Kindly give me a step by step solution.

    Hope to hear from you soon. Thanking you all in advance for your replies.
  2. jcsd
  3. Sep 12, 2013 #2


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    What does it mean to take the partial derivative of a function?
  4. Sep 12, 2013 #3


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    The partial derivative is a "directional derivative" in one of the given coordinate directions. Since you are working in orthogonal coordinates you can ignore the ones not being used ...

    Thus for diff(Z,x) you simply treat y as a constant and then differentiate wrt x as normal:
    diff(Z,x) = 2xy + 2x + 3x^2 +4x^3.

    Now you do diff(Z,y)!
  5. Sep 12, 2013 #4
    Thank you ultrafastped for your reply. Could please explain what you meant by orthogonal corodinates? I would be highly grateful if you can show me step by step. Thank you.

  6. Sep 12, 2013 #5


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    Orthogonal coordinates are a special case of curvilinear coordinates. A set of orthogonal coordinates means that the coordinates that make up the basis of your coordinate system are, well, orthogonal. The most obvious example (and the case in question) is the Cartesian system defined by the ##\left\{i,j,k\right\}## basis.

    Other examples you may have seen include spherical and cylindrical coordinates.
  7. Sep 12, 2013 #6


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    It's easier than it seems!

    If you have the function

    y = 3 + x^2*3 + x^2 + x^3 + x^4 + 5

    Then what is dy/dx?

    Now, I'm sure you can find that, so once you do, since you considered 3 to be a constant, do the same for y when taking the partial derivative dz/dx of the function


    3 is just like y. You treat it like a constant. Then to find dz/dy, just treat x as a constant.
  8. Sep 12, 2013 #7


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    To follow up on orthogonality: orthogonal means independence.

    If two vectors are orthogonal then they are independent and this translates to the situation where changing one vector won't have any impact on changing the other.

    In a non-orthogonal situation, changing one quantity will change another in the general situation.
  9. Sep 13, 2013 #8
    Thank you so much for the reply.
    y = 3 + x^2*3 + x^2 + x^3 + x^4 + 5

    Should the answer be like this:
    dy/dx =6x +2x+3x^2+4x^3 when i consider 3 as constant and is it like 5(number alone without x or y variable) should also be ignored.
  10. Sep 13, 2013 #9


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    You don't need to "consider 3 as constant"- it is constant!

    What they mean is that z= 3xy+ 2x^2- 4xy^2+ x^2y^2, would be thought of as
    1) (differentiating with respect to x) z= Ax+ Bx^2- Cx+ Dx^2, where A= 3y, B= 2, C= 4y^2, and D= y^2 are "constants", so that z_x= A+ 2Bx- C+ 2Dx= 3y+ 4x- y^2+ 2y^2x.

    2) (differentiating with respect to y) z= Ay+ B- Cy^2+ Dy^2, where now A= 3x, B= 2x^2, C= 4x, and D= x^2 are "constants", so that z_y= A- 2Cy+ 2Dy= 3x- 2x^2y+ 2x^2y.
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