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LeoMessi

Quick question to clarify my understanding of how to find the gradient of a function: In practice exam 2b, Problem #1: (x^2)(y^2)-x

  • 2 years ago
  • 2 years ago

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  1. Ackbar
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    all right, to find the gradient of this function, \[f(x,y) = x^2y^2-x\]

    • 2 years ago
  2. Ackbar
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    you find the partial derivative of f with regard to x, and then with regard to y

    • 2 years ago
  3. Ackbar
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    \[<\partial f/\partial x, \partial f/\partial y>\]

    • 2 years ago
  4. LeoMessi
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    all right, so first, partial w/re to x is 2x*y^2-1

    • 2 years ago
  5. LeoMessi
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    and partial w/re to y is 2x^2*y

    • 2 years ago
  6. Ackbar
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    looks good, so format would be: \[<2xy^2-1, 2x^2y>\]

    • 2 years ago
  7. LeoMessi
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    right, excellent, just wanted to make sure my brain was working properly ;)

    • 2 years ago
  8. Ackbar
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    now you can just plug in the point to consider like in the prac. exam problem

    • 2 years ago
  9. sunnyd48
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    idk.

    • 2 years ago
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