Hello Everyone.
I am currently in the process of studying for the FE (Electrical and Comp) Exam.
Trying to learn some concepts I missed in school and was hoping for some advice with the following question:
The Solution is as follows:

So far I've done little more than copy the problem. I'm just not sure where to go from there, if someone could help me understand the solution, I'd be grateful. Thanks and Best Regards.
- Al -

 

What part of the given solution is stumping you? Do you know how to take partial derivatives?

 

Thank you WBahn for the quick reply.
No, I don't believe I do. However, my next step will be to learn the partial derivative procedure, if that's what you suggest.
I'll post back later after trying that.
Best Regards.

 

What background in calculus do you presently have?

 

I took Cal II a few years ago. I do remember briefly going over this subject.... guess it didn't stick .

 

Found this walk through of the solution online.
Thanks for pointing me in the right direction by suggesting Integration by Parts.

 

Where's the x² in the denominator coming from?

 

Correction: I meant to say "Partial Derivatives" not "Integration by parts"

It looks to me as the denominator (x^2) is used to evaluate each part of the equation at x, y, and z

Apparently (a*f)' = a*f' ==> 1/3 a/ax x^3

Then apply the Power Rule: 1/3 3x^3-1 ==> x^2
?

 

No. The differential operator

\(\frac{\partial ^2}{\partial x^2}\) means: take the partial derivative with respect to x, and x only, twice. There is no \(x^2\) in the denominator.

 

I'm not seeing it.

You want the second partial derivative with respect to x of f(x). So take the partial derivative with respect to x twice.

\(
f(x) \; = \; \frac{1}{3}x^3 \, - \, 9y \, + \, 5
\)

So you want

\(
\frac{\partial \, f(x)}{\partial x} \; = \; \frac{\partial \,\frac{1}{3}x^3 \, - \, 9y \, + \, 5}{\partial x}
\)

\(
\frac{\partial \, f(x)}{\partial x} \; = \; \frac{1}{3}\frac{\partial}{\partial x} x^3 \, - \, 9y \frac{\partial}{\partial x} 1 \, + \, 5 \frac{\partial}{\partial x} 1
\)

The last two terms do not depend on x at all, and are therefore zero. The first term reduces

\(
\frac{\partial \, f(x)}{\partial x} \; = \; \frac{1}{3}3x^2 \; = \; x^2
\)

Now take the partial derivative of this with respect to x to get the second partial derivative of f(x) with respect to x and you get

\(
\frac{\partial^2 \, f(x)}{\partial x^2} \; = \; \frac{\partial}{\partial x} x^2 \; = \; 2x
\)

You still need to take the second partials with respect to the other two variables, but when you do you will see that nothing survives.

 

Thank you again - this makes much better sense now.
I hope to get some more help when I get to the circuit analysis section and beyond.