Question

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

Answer 1

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.

Answer 2

do you get this?

Answer 3

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?

Answer 4

yep

Answer 5

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?

Answer 6

I forgot to add at the point (1,1)

Answer 7

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?

Answer 8

X

Answer 9

\[f(x,y)=4x-\frac{ 2y }{ 4x }+2y\]
\[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)\]

Answer 10

can you do this? treat variable y as a constant

Answer 11

Soooo...

Is it 4- 2y/4 +2y ?

Or do the y's go to 0?

Answer 12

Because they are a constant?

Answer 13

nothing goes to zero

Answer 14

y is simply treated as a constant

Answer 15

so it belongs with the number terms

Answer 16

how about if i simplified the function
\[f(x,y)=4x-\frac{ 1 }{ 2 }yx ^{-1}+2y\]

Answer 17

Oh, right! Okay!

Answer 18

So then plug in 1,1?

Answer 19

\[\frac{ \partial f }{ \partial x } = 4\frac{ \partial }{ \partial x }(x)+\frac{ 1 }{ 2 }\frac{ \partial }{ \partial x }(yx ^{-1})+2\frac{ \partial d }{ \partial dx }(y)\]

Answer 20

so the partial derivatie w.r.t to x of x is simply 1 right? we are simply breaking the terms

Answer 21

How did you get 1? Sorry..

Answer 22

if i had y=x what is the derivative dy/dx

Answer 23

1?

Answer 24

yeah you are just using that concept for a function with more than one variable

Answer 25

\[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x-\frac{ 2y }{ 4x }+2y)=4+\frac{ y }{ 2x ^{2} }\]

Answer 26

\[\frac{ \partial f }{ \partial y } = -\frac{ 1 }{ 2x }+2\]

Answer 27

Oh okay! Great! Thank you so much!!

Answer 28

simply plug in (1,1) in each

Answer 29

Right! Thank you!