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malcolm11235

  • 3 years ago

p

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  1. TuringTest
    • 3 years ago
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    you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?

  2. malcolm11235
    • 3 years ago
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    if you show me just one of those partial derivatives i'll know how to find the other

  3. TuringTest
    • 3 years ago
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    do I understand the problem correctly?

  4. TuringTest
    • 3 years ago
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    I don't think I do since it seems to make no sense to me...

  5. malcolm11235
    • 3 years ago
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    the partial derivative of z with repsect to x. del z/ del x or die z / die x . that symbol that is used for partial derivatives that looks like a two kind of

  6. TuringTest
    • 3 years ago
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    ooooh I see, but you are not given any more info besides x=f(xy) ?

  7. TuringTest
    • 3 years ago
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    z=f(xy)

  8. malcolm11235
    • 3 years ago
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    yeah. that's it.

  9. TuringTest
    • 3 years ago
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    well then I suppose all there is to do is apply the chain rule...

  10. malcolm11235
    • 3 years ago
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    so i can start off by going z - f(xy) = 0 then i need to find the partial derivative of x, so f_x then f_y and f_z and del z/ del x = -f_x / f_z

  11. malcolm11235
    • 3 years ago
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    do i let xy =z then use the chain rule?

  12. TuringTest
    • 3 years ago
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    okay I am confused again,\[z=f(xy)\]as in "z equals a function of x times y" correct?

  13. malcolm11235
    • 3 years ago
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    actually no, i know how to do these when there are actual numbers but i'm not quite sure with the f(xy)

  14. malcolm11235
    • 3 years ago
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    yeah

  15. TuringTest
    • 3 years ago
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    oh wait I invented a z, sorry

  16. malcolm11235
    • 3 years ago
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    okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...

  17. TuringTest
    • 3 years ago
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    okay let me rethink this for a sec...

  18. TuringTest
    • 3 years ago
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    oh I see

  19. malcolm11235
    • 3 years ago
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    i know what i want to do but i'm not quite sure how to do it. the only thing throwing me off is f(xy) so if i take the partial derivative of that with respect to x, i hold y constant. if it was a normal expression no problem but since its not i'm a little confused

  20. malcolm11235
    • 3 years ago
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    would it be f ' (x) f(y) ?

  21. TuringTest
    • 3 years ago
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    my reasoning gives\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}f(xy)=\frac{\partial f}{\partial x}\cdot\frac{\partial }{\partial x}(xy)=y\frac{\partial f}{\partial x}=yf_x(xy)\]but the choices in your answer are for f' (whole derivatives on f, not partial)???

  22. malcolm11235
    • 3 years ago
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    that's correct.

  23. malcolm11235
    • 3 years ago
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    correct as in they're not partial derivatives for my choices of answers

  24. TuringTest
    • 3 years ago
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    so I guess then they want\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)\]

  25. TuringTest
    • 3 years ago
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    this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever

  26. malcolm11235
    • 3 years ago
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    okay thanks!

  27. TuringTest
    • 3 years ago
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    welcome!

  28. vf321
    • 3 years ago
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    if the question's asking for partial derivatives then z = f(x*y) makes sense. If the question's asking for the total differential wrt x or y than they could just be asking about the chain rule and z= f(x, y)

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