Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
you want\[\nabla z\over\nabla x\]and\[\nabla z\over\nabla y\]?
if you show me just one of those partial derivatives i'll know how to find the other
do I understand the problem correctly?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I don't think I do since it seems to make no sense to me...
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
ooooh I see, but you are not given any more info besides x=f(xy) ?
z=f(xy)
yeah. that's it.
well then I suppose all there is to do is apply the chain rule...
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
do i let xy =z then use the chain rule?
okay I am confused again,\[z=f(xy)\]as in "z equals a function of x times y" correct?
actually no, i know how to do these when there are actual numbers but i'm not quite sure with the f(xy)
yeah
oh wait I invented a z, sorry
okay but this is a multiple choice question and my options are like f'(x)f(y) or xf '(xy) ect...
okay let me rethink this for a sec...
oh I see
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
would it be f ' (x) f(y) ?
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)???
that's correct.
correct as in they're not partial derivatives for my choices of answers
so I guess then they want\[\frac{\partial }{\partial x}z=\frac{\partial }{\partial x}(xy)f'(xy)=yf'(xy)\]
this would make sense, you can check it with a function like z=sin(xy), z=(xy)^2, whatever
okay thanks!
welcome!
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)

Not the answer you are looking for?

Search for more explanations.

Ask your own question