## malcolm11235 Group Title p one year ago one year ago

1. TuringTest Group Title

you want$\nabla z\over\nabla x$and$\nabla z\over\nabla y$?

2. malcolm11235 Group Title

if you show me just one of those partial derivatives i'll know how to find the other

3. TuringTest Group Title

do I understand the problem correctly?

4. TuringTest Group Title

I don't think I do since it seems to make no sense to me...

5. malcolm11235 Group Title

the partial derivative of z with repsect to x. del z/ del x or die z / die x . that symbol that is used for partial derivatives that looks like a two kind of

6. TuringTest Group Title

ooooh I see, but you are not given any more info besides x=f(xy) ?

7. TuringTest Group Title

z=f(xy)

8. malcolm11235 Group Title

yeah. that's it.

9. TuringTest Group Title

well then I suppose all there is to do is apply the chain rule...

10. malcolm11235 Group Title

so i can start off by going z - f(xy) = 0 then i need to find the partial derivative of x, so f_x then f_y and f_z and del z/ del x = -f_x / f_z

11. malcolm11235 Group Title

do i let xy =z then use the chain rule?

12. TuringTest Group Title

okay I am confused again,$z=f(xy)$as in "z equals a function of x times y" correct?

13. malcolm11235 Group Title

actually no, i know how to do these when there are actual numbers but i'm not quite sure with the f(xy)

14. malcolm11235 Group Title

yeah

15. TuringTest Group Title

oh wait I invented a z, sorry

16. malcolm11235 Group Title

okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...

17. TuringTest Group Title

okay let me rethink this for a sec...

18. TuringTest Group Title

oh I see

19. malcolm11235 Group Title

i know what i want to do but i'm not quite sure how to do it. the only thing throwing me off is f(xy) so if i take the partial derivative of that with respect to x, i hold y constant. if it was a normal expression no problem but since its not i'm a little confused

20. malcolm11235 Group Title

would it be f ' (x) f(y) ?

21. TuringTest Group Title

my reasoning gives$\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}f(xy)=\frac{\partial f}{\partial x}\cdot\frac{\partial }{\partial x}(xy)=y\frac{\partial f}{\partial x}=yf_x(xy)$but the choices in your answer are for f' (whole derivatives on f, not partial)???

22. malcolm11235 Group Title

that's correct.

23. malcolm11235 Group Title

correct as in they're not partial derivatives for my choices of answers

24. TuringTest Group Title

so I guess then they want$\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)$

25. TuringTest Group Title

this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever

26. malcolm11235 Group Title

okay thanks!

27. TuringTest Group Title

welcome!

28. vf321 Group Title

if the question's asking for partial derivatives then z = f(x*y) makes sense. If the question's asking for the total differential wrt x or y than they could just be asking about the chain rule and z= f(x, y)