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

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