# Homework Help: Partial Derivatives

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1. Nov 8, 2016

1. The problem statement, all variables and given/known data

So I know I have to take the derivative with respect to x, then respect to y, then respect to z, but I am not getting the right answer. I know that the answer is 0 and my professor did it with very few steps that I do not understand. Can someone please guide me through it?

2. Nov 8, 2016

### Fightfish

The trick is to recognize that the partial derivative is a linear operator, that is, the partial derivative of a sum is equal to the sum of the partial derivatives of the individual terms:
$$\frac{\partial}{\partial x} \left( f + g \right) = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}$$
Try splitting $f(x,y,z)$ into two parts, take the partial derivatives separately and see where that leads you.

3. Nov 8, 2016

So if I take the partial derivative wrt x and the result has no x term what does that mean?

4. Nov 8, 2016

### Fightfish

Could you provide an explicit example of what you're trying to ask?

5. Nov 8, 2016

So I split the function like you said and I got xln(y100+37z11 / xz rad(y2+1 and when I take the partial derivative of that term wrt x, x is no longer in that term

6. Nov 8, 2016

### Fightfish

The partial derivative means to take all other variables as constants - what happens when you differentiate a constant?

7. Nov 8, 2016

I did take the other variables as constant. I'm saying that by keeping everything constant but x, the x term differentiates to 1 if you look at the function.

8. Nov 8, 2016

### Fightfish

How does it "differentiate to $1$"? Take a closer look at it:
$$\frac{x \ln(y^{100} + 37z^{11})}{x z \sqrt{y^2 +1}} = \frac{\ln(y^{100} + 37z^{11})}{z \sqrt{y^2 +1}}$$
is independent of $x$. So what happens when you take the partial derivative wrt $x$?

9. Nov 8, 2016

Isn't that what you would get if you differentiate with respect to x?

10. Nov 8, 2016

### Fightfish

No, I haven't differentiated - take a closer look: all I did was to divide both the numerator and denominator by $x$.

11. Nov 8, 2016

So that differentiates to 0 then

12. Nov 8, 2016

### Fightfish

Yup. Now look at the second term and see if you notice something similar.

13. Nov 8, 2016

So for the second term, differentiating to z is 0 as well

14. Nov 8, 2016

### Fightfish

Yup. So immediately we conclude that $f_{xyz}$ is indeed $0$.

15. Nov 8, 2016

But we haven't taken the partial derivative wrt y. Don't you have to go in order taking the partial derivative of x, y, then z?

16. Nov 8, 2016

### Fightfish

For a suitably well-behaved function, it doesn't matter - as the hint to the question itself explicitly says. Even if you wanted to "go in order", notice that if I had a function that depends only on $x$ and $y$ but not $z$, when I perform the partial derivatives wrt $x$ and $y$, I will still end up with a result that is independent of $z$ i.e. I won't have a $z$ suddenly appearing in my expression. So in the end, when I take the partial derivative wrt $z$, it will still be $0$.

17. Nov 8, 2016

I think my confusion lies in splitting the function into two parts. So for the first part of the function differentiating wrt x gives 0 but that's not the case for the second part of the function. Why do I ignore that second part for x?

18. Nov 8, 2016

Like, wouldn't I differentiate the entire function in terms of x?

19. Nov 8, 2016

### Fightfish

What we are actually doing by splitting is to evaluate separately and then add the result together.
Suppose we split a function $f(x,y,z)$ into two parts, say
$$f(x,y,z) = g(x,y,z) + h(x,y,z)$$
Then, because partial derivation is linear, we have
$$f_{xyz} = g_{xyz} + h_{xyz}$$
so we can find $g_{xyz}$ and $h_{xyz}$ separately and combine the results to get $f_{xyz}$.

20. Nov 8, 2016