## ranyai12 Group Title 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

1. jtvatsim Group Title

You mean $f_x, f_y, \enspace \text{and}\enspace f_{xxy}$ right?

2. jtvatsim Group Title

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.

3. jtvatsim Group Title

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).

4. jtvatsim Group Title

Does that make sense to you? :)

5. ranyai12 Group Title

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

6. ranyai12 Group Title

I figured out my mistake but whats fxxy?

7. jtvatsim Group Title

OK, so for f_y we will get the following

8. jtvatsim Group Title

brb

9. ranyai12 Group Title

i solved for f_y but I cant seem to figure out what f_xxy is supposed to be

10. jtvatsim Group Title

kk

11. jtvatsim Group Title

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.

12. jtvatsim Group Title

It's like doing the third derivative, but you take the derivative with respect to different variable each time

13. jtvatsim Group Title

$f(x,y)=5(x^ 6 )y+4\sin(xy)$ $f_x = 30yx^5 + 4y \cos(xy)$ $f_{xx} = 150yx^4 - 4y^2 \sin(xy)$ $f_{xxy}=150x^4-4[y^2\cdot y \cos(xy) + \sin(xy)\cdot 2y]$

14. jtvatsim Group Title

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.

15. ranyai12 Group Title

im sorry but can you explain that a little better im still confused

16. ranyai12 Group Title

@jtvatsim

17. jtvatsim Group Title

18. ranyai12 Group Title

thank you but the answer is wrong :(

19. ranyai12 Group Title

@agent0smith d you have any idea on how to solve for the last part?

20. zepdrix Group Title

Were you able to solve the first two parts? :o

21. ranyai12 Group Title

yes i did but i cant seem to figure out the last part

22. zepdrix Group Title

Yah Jvatsim made a tiny mistake on his last derivative, let's continue from $$f_{xx}$$.

23. ranyai12 Group Title

ok thank you

24. zepdrix Group Title

$\large f_{xx} =\color{royalblue}{150x^4}y - 4y^2 \sin(\color{royalblue}{x}y)$ 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? $\large 150x^4$

25. zepdrix Group Title

For the next part, we have two functions of y, so we have to apply the product rule.

26. ranyai12 Group Title

ok but how would we do that

27. zepdrix Group Title

$\large \color{orangered}{(-4y^2)'}\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}$ 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.

28. zepdrix Group Title

$\large (-8y)\sin(xy)-4y^2\color{orangered}{\left[\sin(xy)\right]'}$Understand the first term ok? I guess remembering how to setup product rule is pretty important for this :\

29. ranyai12 Group Title

yea i do thanks

30. ranyai12 Group Title

for the second part would you consider x as a constant

31. zepdrix Group Title

Yes c: You'll end up applying the chain rule after you take the derivative of the sine portion.

32. zepdrix Group Title

So yah the x being a constant is important to remember c:

33. zepdrix Group Title

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

34. ranyai12 Group Title

so the second part would be xcos(xy)?

35. ranyai12 Group Title

(−8y)sin(xy)−4xy^2[cos(xy)]

36. zepdrix Group Title

yes good.

37. ranyai12 Group Title

so would that be the answer?

38. zepdrix Group Title

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

39. ranyai12 Group Title

haha i ddnt forget thank you sooo much!