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ranyai12 Group Title

Can someone please help?! For the function f(x,y)=5(x^6)y+4sin(xy), calculate the indicated partial derivatives f(x)= f(y)= f(xxy)=

  • one year ago
  • one year ago

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  1. jtvatsim Group Title
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    You mean \[f_x, f_y, \enspace \text{and}\enspace f_{xxy}\] right?

    • one year ago
  2. jtvatsim Group Title
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    So, for \[f_x\] treat y like a constant and take the derivative as usual. In other words, pretend that y is the number 2 or something and do the derivative of x.

    • one year ago
  3. jtvatsim Group Title
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    So for, \[f(x,y)=5(x^6)y+4\sin(xy)\] Treat y like any old number and we get that \[f_x=5y\cdot 6x^5 + 4\cdot y \cos(xy)=30x^5y+4y \cos(xy)\] Notice how we used the chain rule for sin(xy) and thought of it just like the derivative of sin(2x) = 2cos(2x).

    • one year ago
  4. jtvatsim Group Title
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    Does that make sense to you? :)

    • one year ago
  5. ranyai12 Group Title
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    ok so i think i got it but when I attempted for f_y i got 5x^6-4xsin(xy) and it was wrong can you lease tell me what im doing wrong

    • one year ago
  6. ranyai12 Group Title
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    I figured out my mistake but whats fxxy?

    • one year ago
  7. jtvatsim Group Title
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    OK, so for f_y we will get the following

    • one year ago
  8. jtvatsim Group Title
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    brb

    • one year ago
  9. ranyai12 Group Title
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    i solved for f_y but I cant seem to figure out what f_xxy is supposed to be

    • one year ago
  10. jtvatsim Group Title
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    kk

    • one year ago
  11. jtvatsim Group Title
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    So for f_xxy, first do f_x you will get some equation. Then take the equation you get and do f_x again you will get another equation. Last, take the equation and do f_y and thats the answer.

    • one year ago
  12. jtvatsim Group Title
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    It's like doing the third derivative, but you take the derivative with respect to different variable each time

    • one year ago
  13. jtvatsim Group Title
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    \[f(x,y)=5(x^ 6 )y+4\sin(xy)\] \[f_x = 30yx^5 + 4y \cos(xy)\] \[f_{xx} = 150yx^4 - 4y^2 \sin(xy)\] \[f_{xxy}=150x^4-4[y^2\cdot y \cos(xy) + \sin(xy)\cdot 2y]\]

    • one year ago
  14. jtvatsim Group Title
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    you must use a product rule for the last part, think of having 4y^2 times sin(2y) you would have to use the product rule.

    • one year ago
  15. ranyai12 Group Title
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    im sorry but can you explain that a little better im still confused

    • one year ago
  16. ranyai12 Group Title
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    @jtvatsim

    • one year ago
  17. jtvatsim Group Title
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    http://www.youtube.com/watch?v=2ZwdHLbmFJs fast forward to 22:15

    • one year ago
  18. ranyai12 Group Title
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    thank you but the answer is wrong :(

    • one year ago
  19. ranyai12 Group Title
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    @agent0smith d you have any idea on how to solve for the last part?

    • one year ago
  20. zepdrix Group Title
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    Were you able to solve the first two parts? :o

    • one year ago
  21. ranyai12 Group Title
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    yes i did but i cant seem to figure out the last part

    • one year ago
  22. zepdrix Group Title
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    Yah Jvatsim made a tiny mistake on his last derivative, let's continue from \(f_{xx}\).

    • one year ago
  23. ranyai12 Group Title
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    ok thank you

    • one year ago
  24. zepdrix Group Title
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    \[\large f_{xx} =\color{royalblue}{150x^4}y - 4y^2 \sin(\color{royalblue}{x}y)\] We're going to be taking the derivative with respect to `y` for our last step, so we can think of these blue values as constants. The first part shouldn't be too bad, ~A constant times y, taking the derivative with respect to y will leave us with just the constant yes? \[\large 150x^4\]

    • one year ago
  25. zepdrix Group Title
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    For the next part, we have two functions of `y`, so we have to apply the product rule.

    • one year ago
  26. ranyai12 Group Title
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    ok but how would we do that

    • one year ago
  27. zepdrix Group Title
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    \[\large \color{orangered}{(-4y^2)'}\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}\] Sorry for the delay, got sidetracked. So here is how we setup the product rule, We need to take the derivative of the orange terms. And we're differentiating these orange parts only with respect to y.

    • one year ago
  28. zepdrix Group Title
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    \[\large (-8y)\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}\]Understand the first term ok? I guess remembering how to setup product rule is pretty important for this :\

    • one year ago
  29. ranyai12 Group Title
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    yea i do thanks

    • one year ago
  30. ranyai12 Group Title
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    for the second part would you consider x as a constant

    • one year ago
  31. zepdrix Group Title
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    Yes c: You'll end up applying the chain rule after you take the derivative of the sine portion.

    • one year ago
  32. zepdrix Group Title
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    So yah the x being a constant is important to remember c:

    • one year ago
  33. zepdrix Group Title
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    In the previous partials you did, the y was the constant, so an extra y kept popping out each time you took a derivative. The opposite variable will pop out this time

    • one year ago
  34. ranyai12 Group Title
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    so the second part would be xcos(xy)?

    • one year ago
  35. ranyai12 Group Title
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    (−8y)sin(xy)−4xy^2[cos(xy)]

    • one year ago
  36. zepdrix Group Title
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    yes good.

    • one year ago
  37. ranyai12 Group Title
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    so would that be the answer?

    • one year ago
  38. zepdrix Group Title
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    Yes, you have 3 parts I believe. The first part, and then the other two parts which came from the product rule.\[\large f_{xxy}=150x^4-8ysin(xy)-4xy^2\cos(xy)\] We were working on each part separately, hopefully you didn't lose track of that first part :) heh

    • one year ago
  39. ranyai12 Group Title
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    haha i ddnt forget thank you sooo much!

    • one year ago
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