Operator help

- AmTran_Bus

Operator help

- jamiebookeater

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- AmTran_Bus

##### 1 Attachment

- beginnersmind

So what's the question? Calculate A*f(x,y,x) ?

- AmTran_Bus

Exactly.

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## More answers

- 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

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

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

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

Can you take a quick guess first?

- AmTran_Bus

Sure. 6xy^2z^4 + 2x^3z^4+12x^3y^2z^2

- beginnersmind

One sec

- Jhannybean

begin by writing your first partials \[f' =\langle 3x^2y^2z^4~,~ 2x^3yz^4~,~ 4x^3y^2z^3\rangle \]

- Jhannybean

Then simply take their derivative in component form :)

- beginnersmind

Ok, so that's actually the correct answer. I guess the question is which point do you have trouble with?

- 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

(just how I would approach it!)

- AmTran_Bus

Ohh ok. Thanks so much @Jhannybean

- Jhannybean

but yeah, along with what @beginnersmind stated, where were ya having difficulty?

- AmTran_Bus

Ive never seen this stuff before and its on a pchem homework problem

- AmTran_Bus

like, i understand taking the derivative, but where does the 6, 2, and 12 come from @Jhannybean

- AmTran_Bus

- beginnersmind

Well, you tell me, it's you who calculated it.

- beginnersmind

(it is correct though)

- AmTran_Bus

No no, it was the soln man that did!!!!

- 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

If you don't know what it means, that's ok too, but I need to know where to start my explanation.

- 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

The 6, 2 and 12 come from deriving each component based on what variable you are taking the partial derivative of.

- 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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