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ranyai12

  • 2 years 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
    • 2 years ago
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    You mean \[f_x, f_y, \enspace \text{and}\enspace f_{xxy}\] right?

  2. jtvatsim
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Does that make sense to you? :)

  5. ranyai12
    • 2 years 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
    • 2 years ago
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    I figured out my mistake but whats fxxy?

  7. jtvatsim
    • 2 years ago
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    OK, so for f_y we will get the following

  8. jtvatsim
    • 2 years ago
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    brb

  9. ranyai12
    • 2 years 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
    • 2 years ago
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    kk

  11. jtvatsim
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    im sorry but can you explain that a little better im still confused

  16. ranyai12
    • 2 years ago
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    @jtvatsim

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

  18. ranyai12
    • 2 years ago
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    thank you but the answer is wrong :(

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

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

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

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

  23. ranyai12
    • 2 years ago
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    ok thank you

  24. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    ok but how would we do that

  27. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yea i do thanks

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

  31. zepdrix
    • 2 years 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
    • 2 years ago
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    So yah the x being a constant is important to remember c:

  33. zepdrix
    • 2 years 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
    • 2 years ago
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    so the second part would be xcos(xy)?

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

  36. zepdrix
    • 2 years ago
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    yes good.

  37. ranyai12
    • 2 years ago
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    so would that be the answer?

  38. zepdrix
    • 2 years 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
    • 2 years ago
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    haha i ddnt forget thank you sooo much!

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