Find the first partial derivatives of f(x, y) = sin(x - y) at the point (7, 7).

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Find the first partial derivatives of f(x, y) = sin(x - y) at the point (7, 7).

Mathematics
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Using the chain rule gives\[\frac{ \partial f }{ \partial x } = \cos \left( x-y \right)\]\[\frac{ \partial f }{ \partial y } = -\cos \left( x-y \right)\]Evaluate using x=7 and y=7.
do you get this?
Thank you for setting it up for me! cos(7-7)= cos(0) = 1 -cos(7-7)= -cos(0) = -1 Is that all you have to do?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yep
Thank you!! I have one more question similar to it! Find the first partial derivatives of f(x,y) = 4x-2y/4x+2y Can you help me set this up?
I forgot to add at the point (1,1)
well when you do partial, you must take the derivative of what you want and then treat all other vairables as constants. so what do you want to differentitate with respect to first? x or y?
X
\[f(x,y)=4x-\frac{ 2y }{ 4x }+2y\] \[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)\]
can you do this? treat variable y as a constant
Soooo... Is it 4- 2y/4 +2y ? Or do the y's go to 0?
Because they are a constant?
nothing goes to zero
y is simply treated as a constant
so it belongs with the number terms
how about if i simplified the function \[f(x,y)=4x-\frac{ 1 }{ 2 }yx ^{-1}+2y\]
Oh, right! Okay!
So then plug in 1,1?
\[\frac{ \partial f }{ \partial x } = 4\frac{ \partial }{ \partial x }(x)+\frac{ 1 }{ 2 }\frac{ \partial }{ \partial x }(yx ^{-1})+2\frac{ \partial d }{ \partial dx }(y)\]
so the partial derivatie w.r.t to x of x is simply 1 right? we are simply breaking the terms
How did you get 1? Sorry..
if i had y=x what is the derivative dy/dx
1?
yeah you are just using that concept for a function with more than one variable
\[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)=4+\frac{ y }{ 2x ^{2} }\]
\[\frac{ \partial f }{ \partial y } = -\frac{ 1 }{ 2x }+2\]
Oh okay! Great! Thank you so much!!
simply plug in (1,1) in each
Right! Thank you!

Not the answer you are looking for?

Search for more explanations.

Ask your own question