AmTran_Bus
  • AmTran_Bus
Operator help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AmTran_Bus
  • AmTran_Bus
beginnersmind
  • beginnersmind
So what's the question? Calculate A*f(x,y,x) ?
AmTran_Bus
  • AmTran_Bus
Exactly.

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More answers

beginnersmind
  • beginnersmind
Ok, 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
  • AmTran_Bus
Thanks 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
  • beginnersmind
I 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
  • AmTran_Bus
Yea, 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
  • beginnersmind
Can you take a quick guess first?
AmTran_Bus
  • AmTran_Bus
Sure. 6xy^2z^4 + 2x^3z^4+12x^3y^2z^2
beginnersmind
  • beginnersmind
One sec
Jhannybean
  • Jhannybean
begin by writing your first partials \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle \]
Jhannybean
  • Jhannybean
Then simply take their derivative in component form :)
beginnersmind
  • beginnersmind
Ok, so that's actually the correct answer. I guess the question is which point do you have trouble with?
Jhannybean
  • Jhannybean
I wrote it in component form so you could see their derivatives, it's asking for the additive operative, so just add them instead.
Jhannybean
  • Jhannybean
(just how I would approach it!)
AmTran_Bus
  • AmTran_Bus
Ohh ok. Thanks so much @Jhannybean
Jhannybean
  • Jhannybean
but yeah, along with what @beginnersmind stated, where were ya having difficulty?
AmTran_Bus
  • AmTran_Bus
Ive never seen this stuff before and its on a pchem homework problem
AmTran_Bus
  • AmTran_Bus
like, i understand taking the derivative, but where does the 6, 2, and 12 come from @Jhannybean
AmTran_Bus
  • AmTran_Bus
or @beginnersmind
beginnersmind
  • beginnersmind
Well, you tell me, it's you who calculated it.
beginnersmind
  • beginnersmind
(it is correct though)
AmTran_Bus
  • AmTran_Bus
No no, it was the soln man that did!!!!
beginnersmind
  • beginnersmind
Oh. 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
  • beginnersmind
If you don't know what it means, that's ok too, but I need to know where to start my explanation.
Jhannybean
  • Jhannybean
hmm... \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle\]\[f''=\langle \color{red}{6x}y^2z^4~,~ \color{red}{2}x^3z^4~,~ \color{red}{12z^2}x^3y^2\rangle \]
Jhannybean
  • Jhannybean
The 6, 2 and 12 come from deriving each component based on what variable you are taking the partial derivative of.
Jhannybean
  • Jhannybean
do 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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