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 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)=
 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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jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1You mean \[f_x, f_y, \enspace \text{and}\enspace f_{xxy}\] right?

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1So, 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.

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1So 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).

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1Does that make sense to you? :)

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0ok so i think i got it but when I attempted for f_y i got 5x^64xsin(xy) and it was wrong can you lease tell me what im doing wrong

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0I figured out my mistake but whats fxxy?

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1OK, so for f_y we will get the following

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0i solved for f_y but I cant seem to figure out what f_xxy is supposed to be

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1So 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.

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1It's like doing the third derivative, but you take the derivative with respect to different variable each time

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1\[f(x,y)=5(x^ 6 )y+4\sin(xy)\] \[f_x = 30yx^5 + 4y \cos(xy)\] \[f_{xx} = 150yx^4  4y^2 \sin(xy)\] \[f_{xxy}=150x^44[y^2\cdot y \cos(xy) + \sin(xy)\cdot 2y]\]

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1you 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.

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0im sorry but can you explain that a little better im still confused

jtvatsim
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.youtube.com/watch?v=2ZwdHLbmFJs fast forward to 22:15

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0thank you but the answer is wrong :(

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0@agent0smith d you have any idea on how to solve for the last part?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Were you able to solve the first two parts? :o

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0yes i did but i cant seem to figure out the last part

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yah Jvatsim made a tiny mistake on his last derivative, let's continue from \(f_{xx}\).

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large f_{xx} =\color{royalblue}{150x^4}y  4y^2 \sin(\color{royalblue}{x}y)\] 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? \[\large 150x^4\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1For the next part, we have two functions of `y`, so we have to apply the product rule.

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0ok but how would we do that

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \color{orangered}{(4y^2)'}\sin(xy)4y^2\color{orangered}{\left[\sin(xy)\right]'}\] 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.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large (8y)\sin(xy)4y^2\color{orangered}{\left[\sin(xy)\right]'}\]Understand the first term ok? I guess remembering how to setup product rule is pretty important for this :\

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0for the second part would you consider x as a constant

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes c: You'll end up applying the chain rule after you take the derivative of the sine portion.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So yah the x being a constant is important to remember c:

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1In 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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0so the second part would be xcos(xy)?

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0(−8y)sin(xy)−4xy^2[cos(xy)]

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0so would that be the answer?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, you have 3 parts I believe. The first part, and then the other two parts which came from the product rule.\[\large f_{xxy}=150x^48ysin(xy)4xy^2\cos(xy)\] We were working on each part separately, hopefully you didn't lose track of that first part :) heh

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0haha i ddnt forget thank you sooo much!
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