credmond
  • credmond
Find the first partial derivatives of f(x, y) = sin(x - y) at the point (7, 7).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
anonymous
  • anonymous
do you get this?
credmond
  • credmond
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?

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

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