# Question on Partial Derivative.

1. Oct 10, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
The function given is
(1+xz)^(1/2) + (1-xy)^(1/2)
I have to take the partial derivative with respect to x, y, and z. The question says Choose the order wisely. I don't understand what it means? How could I choose the order badly? Can anyone skilled in explaining math to simpletons explain this to me?

2. Relevant equations

3. The attempt at a solution

2. Oct 10, 2013

### Staff: Mentor

From what you wrote it isn't clear to me what you are asked to do. You can take the (first) partial with respect to x or the (first) partial with respect to y or the (first) partial with respect to z.

I suspect you are being asked to take the third partial of your function with respect to x, y, and z, in some order. Is that it?

3. Oct 10, 2013

### PsychonautQQ

Yes, I am being asked to take the third partial of my function with respect to x, y, and z in some order. And apparently I get to pick the order and the order matters? Sorry for not being clear this lingo is a bit new to me.

4. Oct 10, 2013

### DimReg

I don't think the question is warning you about a "wrong" way to do it. Rather, it seems to be warning against a harder way to do it. The order doesn't matter for getting the right answer, but it does make getting the answer easier.

Consider the first term. If you first take the derivative with respect to x or z you'll have to compute the derivative. But what happens when you take the derivative with respect to y?

5. Oct 10, 2013

### PsychonautQQ

So basically since both terms have an x in it, don't take the derivative of x first?

6. Oct 10, 2013

### PsychonautQQ

i'm confused... So lets say I take the partial derivative with respect to y first.. the first term will go to zero and then using the chain rule on the second I will get -x / 2(1-(xy))^(1/2)... so then I still have to take the derivative with respect to x and z.. If I take the derivative with respect to z won't the whole thing just go to zero? I suck at derivatives but I really need to get this question right because I'm part of a group and don't wanna screw my groupmates over ;-/

7. Oct 10, 2013

### DimReg

Sounds like you have the right idea. The derivative is linear, which means you can calculate it term by term. Just relax and think about it:

You are taking a derivative with respect to y. The first term does not depend on y. Therefore the first term will contribute nothing to your final answer.

You are also taking a derivative with respect to z. The second term does not depend on z. Therefore the second term will contribute nothing to your final answer as well.

So yes, the answer will just be zero.

One final point: working in a group is great, but try to solve every problem on your own. You'll learn a lot more that way, and sometimes you'll catch mistakes that your friends made. If you're still struggling with derivatives, even though you've moved onto multi-variable calc, then you probably need more practice in general.