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credmond
 one year ago
Find the first partial derivatives of f(x, y) = sin(x  y) at the point (7, 7).
credmond
 one year ago
Find the first partial derivatives of f(x, y) = sin(x  y) at the point (7, 7).

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using the chain rule gives\[\frac{ \partial f }{ \partial x } = \cos \left( xy \right)\]\[\frac{ \partial f }{ \partial y } = \cos \left( xy \right)\]Evaluate using x=7 and y=7.

credmond
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for setting it up for me! cos(77)= cos(0) = 1 cos(77)= cos(0) = 1 Is that all you have to do?

credmond
 one year ago
Best ResponseYou've already chosen the best response.0Thank you!! I have one more question similar to it! Find the first partial derivatives of f(x,y) = 4x2y/4x+2y Can you help me set this up?

credmond
 one year ago
Best ResponseYou've already chosen the best response.0I forgot to add at the point (1,1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x,y)=4x\frac{ 2y }{ 4x }+2y\] \[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x\frac{ 2y }{ 4x }+2y)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you do this? treat variable y as a constant

credmond
 one year ago
Best ResponseYou've already chosen the best response.0Soooo... Is it 4 2y/4 +2y ? Or do the y's go to 0?

credmond
 one year ago
Best ResponseYou've already chosen the best response.0Because they are a constant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nothing goes to zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y is simply treated as a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it belongs with the number terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about if i simplified the function \[f(x,y)=4x\frac{ 1 }{ 2 }yx ^{1}+2y\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \partial f }{ \partial x } = 4\frac{ \partial }{ \partial x }(x)+\frac{ 1 }{ 2 }\frac{ \partial }{ \partial x }(yx ^{1})+2\frac{ \partial d }{ \partial dx }(y)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the partial derivatie w.r.t to x of x is simply 1 right? we are simply breaking the terms

credmond
 one year ago
Best ResponseYou've already chosen the best response.0How did you get 1? Sorry..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if i had y=x what is the derivative dy/dx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah you are just using that concept for a function with more than one variable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \partial f }{ \partial x } = \frac{ \partial }{ \partial x }(4x\frac{ 2y }{ 4x }+2y)=4+\frac{ y }{ 2x ^{2} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \partial f }{ \partial y } = \frac{ 1 }{ 2x }+2\]

credmond
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay! Great! Thank you so much!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0simply plug in (1,1) in each
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