A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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)=
 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)=

This Question is Closed

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

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

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

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

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

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

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

ranyai12
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Were you able to solve the first two parts? :o

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

zepdrix
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0ok but how would we do that

zepdrix
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0for the second part would you consider x as a constant

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

zepdrix
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0so the second part would be xcos(xy)?

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

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

zepdrix
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0haha i ddnt forget thank you sooo much!
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.