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AmTran_Bus
 one year ago
Operator help
AmTran_Bus
 one year ago
Operator help

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beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3So what's the question? Calculate A*f(x,y,x) ?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Ok, so dw:1441654151789:dw That's what this operator is. It takes partial derivatives of a function and adds them together. Do you understand how to take partial derivatives?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much. I am really not all that sure. I know you start with the x and then go with the y and then the z. Like, the x only has the far left to deal with, right?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3I mean, if you are given a function f(x,y,z) = x*y + 2z, can you calculate its partial derivative with respect to x? If not, you need to review that first.

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Yea, so never had that class. In P chem, thrown at us blindly. So I know the answer, but I would really just app. if you could show me the steps.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Can you take a quick guess first?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Sure. 6xy^2z^4 + 2x^3z^4+12x^3y^2z^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0begin by writing your first partials \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then simply take their derivative in component form :)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Ok, so that's actually the correct answer. I guess the question is which point do you have trouble with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wrote it in component form so you could see their derivatives, it's asking for the additive operative, so just add them instead.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(just how I would approach it!)

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Ohh ok. Thanks so much @Jhannybean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but yeah, along with what @beginnersmind stated, where were ya having difficulty?

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Ive never seen this stuff before and its on a pchem homework problem

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0like, i understand taking the derivative, but where does the 6, 2, and 12 come from @Jhannybean

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Well, you tell me, it's you who calculated it.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3(it is correct though)

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0No no, it was the soln man that did!!!!

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Oh. So the question remains, do you know what a partial derivative is and how you calculate it? If you do please calculate dw:1441655336477:dw

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3If you don't know what it means, that's ok too, but I need to know where to start my explanation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm... \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle\]\[f''=\langle \color{red}{6x}y^2z^4~,~ \color{red}{2}x^3z^4~,~ \color{red}{12z^2}x^3y^2\rangle \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 6, 2 and 12 come from deriving each component based on what variable you are taking the partial derivative of.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you remember basic derivatives? Like.... taking the second derivative of \(f(x) = 2x^4\) would be \(f'(x) = (2 \cdot 4)x^3 = 8x^3 \rightarrow f''(x) = (8\cdot 3)x^2 = 16x^2 \) But now you're doing it with each separate component, \(x\) , \(y\) AND \(z\), hence the gradient.
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