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ranyai12

  • one year ago

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)=

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  1. jtvatsim
    • one year ago
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    You mean \[f_x, f_y, \enspace \text{and}\enspace f_{xxy}\] right?

  2. jtvatsim
    • one year ago
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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.

  3. jtvatsim
    • one year ago
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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).

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

  5. ranyai12
    • one year ago
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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

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

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

  8. jtvatsim
    • one year ago
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    brb

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

  10. jtvatsim
    • one year ago
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    kk

  11. jtvatsim
    • one year ago
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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.

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

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

  14. jtvatsim
    • one year ago
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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.

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

  16. ranyai12
    • one year ago
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    @jtvatsim

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

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

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

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

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

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

  23. ranyai12
    • one year ago
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    ok thank you

  24. zepdrix
    • one year ago
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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? \[\large 150x^4\]

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

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

  27. zepdrix
    • one year ago
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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.

  28. zepdrix
    • one year ago
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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 :\

  29. ranyai12
    • one year ago
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    yea i do thanks

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

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

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

  33. zepdrix
    • one year ago
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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

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

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

  36. zepdrix
    • one year ago
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    yes good.

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

  38. zepdrix
    • one year ago
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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

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

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