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

jtvatsimBest ResponseYou've already chosen the best response.1
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

jtvatsimBest ResponseYou've already chosen the best response.1
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

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

ranyai12Best ResponseYou've already chosen the best response.0
ok 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
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
I figured out my mistake but whats fxxy?
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
i solved for f_y but I cant seem to figure out what f_xxy is supposed to be
 one year ago

jtvatsimBest ResponseYou've already chosen the best response.1
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

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

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

jtvatsimBest ResponseYou've already chosen the best response.1
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

ranyai12Best ResponseYou've already chosen the best response.0
im sorry but can you explain that a little better im still confused
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
thank you but the answer is wrong :(
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
@agent0smith d you have any idea on how to solve for the last part?
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
yes i did but i cant seem to figure out the last part
 one year ago

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

zepdrixBest 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? \[\large 150x^4\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
For the next part, we have two functions of `y`, so we have to apply the product rule.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok but how would we do that
 one year ago

zepdrixBest 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.
 one year ago

zepdrixBest 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 :\
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
for the second part would you consider x as a constant
 one year ago

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

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

zepdrixBest ResponseYou've already chosen the best response.1
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

ranyai12Best ResponseYou've already chosen the best response.0
so the second part would be xcos(xy)?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
(−8y)sin(xy)−4xy^2[cos(xy)]
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
so would that be the answer?
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
haha i ddnt forget thank you sooo much!
 one year ago
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