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## AmTran_Bus one year ago Operator help

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1. AmTran_Bus

2. beginnersmind

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

3. AmTran_Bus

Exactly.

4. 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?

5. 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?

6. 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.

7. 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.

8. beginnersmind

Can you take a quick guess first?

9. AmTran_Bus

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

10. beginnersmind

One sec

11. Jhannybean

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

12. Jhannybean

Then simply take their derivative in component form :)

13. beginnersmind

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

14. 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.

15. Jhannybean

(just how I would approach it!)

16. AmTran_Bus

Ohh ok. Thanks so much @Jhannybean

17. Jhannybean

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

18. AmTran_Bus

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

19. AmTran_Bus

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

20. AmTran_Bus

or @beginnersmind

21. beginnersmind

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

22. beginnersmind

(it is correct though)

23. AmTran_Bus

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

24. 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|

25. beginnersmind

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

26. 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$

27. Jhannybean

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

28. 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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