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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)=

Mathematics
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You mean \[f_x, f_y, \enspace \text{and}\enspace f_{xxy}\] right?
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.
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).

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Other answers:

Does that make sense to you? :)
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
I figured out my mistake but whats fxxy?
OK, so for f_y we will get the following
brb
i solved for f_y but I cant seem to figure out what f_xxy is supposed to be
kk
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.
It's like doing the third derivative, but you take the derivative with respect to different variable each time
\[f(x,y)=5(x^ 6 )y+4\sin(xy)\] \[f_x = 30yx^5 + 4y \cos(xy)\] \[f_{xx} = 150yx^4 - 4y^2 \sin(xy)\] \[f_{xxy}=150x^4-4[y^2\cdot y \cos(xy) + \sin(xy)\cdot 2y]\]
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.
im sorry but can you explain that a little better im still confused
http://www.youtube.com/watch?v=2ZwdHLbmFJs fast forward to 22:15
thank you but the answer is wrong :(
@agent0smith d you have any idea on how to solve for the last part?
Were you able to solve the first two parts? :o
yes i did but i cant seem to figure out the last part
Yah Jvatsim made a tiny mistake on his last derivative, let's continue from \(f_{xx}\).
ok thank you
\[\large f_{xx} =\color{royalblue}{150x^4}y - 4y^2 \sin(\color{royalblue}{x}y)\] 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? \[\large 150x^4\]
For the next part, we have two functions of `y`, so we have to apply the product rule.
ok but how would we do that
\[\large \color{orangered}{(-4y^2)'}\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}\] 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.
\[\large (-8y)\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}\]Understand the first term ok? I guess remembering how to setup product rule is pretty important for this :\
yea i do thanks
for the second part would you consider x as a constant
Yes c: You'll end up applying the chain rule after you take the derivative of the sine portion.
So yah the x being a constant is important to remember c:
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
so the second part would be xcos(xy)?
(−8y)sin(xy)−4xy^2[cos(xy)]
yes good.
so would that be the answer?
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
haha i ddnt forget thank you sooo much!

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