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credmond

  • one year ago

Find the first partial derivatives of f(x, y) = sin(x - y) at the point (7, 7).

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  1. anonymous
    • one year ago
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    Using the chain rule gives\[\frac{ \partial f }{ \partial x } = \cos \left( x-y \right)\]\[\frac{ \partial f }{ \partial y } = -\cos \left( x-y \right)\]Evaluate using x=7 and y=7.

  2. anonymous
    • one year ago
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    do you get this?

  3. credmond
    • one year ago
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    Thank you for setting it up for me! cos(7-7)= cos(0) = 1 -cos(7-7)= -cos(0) = -1 Is that all you have to do?

  4. anonymous
    • one year ago
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    yep

  5. credmond
    • one year ago
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    Thank you!! I have one more question similar to it! Find the first partial derivatives of f(x,y) = 4x-2y/4x+2y Can you help me set this up?

  6. credmond
    • one year ago
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    I forgot to add at the point (1,1)

  7. anonymous
    • one year ago
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    well when you do partial, you must take the derivative of what you want and then treat all other vairables as constants. so what do you want to differentitate with respect to first? x or y?

  8. credmond
    • one year ago
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    X

  9. anonymous
    • one year ago
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    \[f(x,y)=4x-\frac{ 2y }{ 4x }+2y\] \[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)\]

  10. anonymous
    • one year ago
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    can you do this? treat variable y as a constant

  11. credmond
    • one year ago
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    Soooo... Is it 4- 2y/4 +2y ? Or do the y's go to 0?

  12. credmond
    • one year ago
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    Because they are a constant?

  13. anonymous
    • one year ago
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    nothing goes to zero

  14. anonymous
    • one year ago
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    y is simply treated as a constant

  15. anonymous
    • one year ago
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    so it belongs with the number terms

  16. anonymous
    • one year ago
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    how about if i simplified the function \[f(x,y)=4x-\frac{ 1 }{ 2 }yx ^{-1}+2y\]

  17. credmond
    • one year ago
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    Oh, right! Okay!

  18. credmond
    • one year ago
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    So then plug in 1,1?

  19. anonymous
    • one year ago
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    \[\frac{ \partial f }{ \partial x } = 4\frac{ \partial }{ \partial x }(x)+\frac{ 1 }{ 2 }\frac{ \partial }{ \partial x }(yx ^{-1})+2\frac{ \partial d }{ \partial dx }(y)\]

  20. anonymous
    • one year ago
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    so the partial derivatie w.r.t to x of x is simply 1 right? we are simply breaking the terms

  21. credmond
    • one year ago
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    How did you get 1? Sorry..

  22. anonymous
    • one year ago
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    if i had y=x what is the derivative dy/dx

  23. credmond
    • one year ago
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    1?

  24. anonymous
    • one year ago
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    yeah you are just using that concept for a function with more than one variable

  25. anonymous
    • one year ago
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    \[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)=4+\frac{ y }{ 2x ^{2} }\]

  26. anonymous
    • one year ago
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    \[\frac{ \partial f }{ \partial y } = -\frac{ 1 }{ 2x }+2\]

  27. credmond
    • one year ago
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    Oh okay! Great! Thank you so much!!

  28. anonymous
    • one year ago
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    simply plug in (1,1) in each

  29. credmond
    • one year ago
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    Right! Thank you!

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