Partial Derivatives: Solve Homework Quickly

[Adeel Ahmad]

Homework Statement

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?

 

[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:
[tex]\frac{\partial}{\partial x} \left( f + g \right) = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} [/tex]
Try splitting ##f(x,y,z)## into two parts, take the partial derivatives separately and see where that leads you.

 

[Adeel Ahmad]

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

 

[Fightfish]

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

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

 

[Adeel Ahmad]

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

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

 

[Fightfish]

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

 

[Adeel Ahmad]

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

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.

 

[Fightfish]

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.

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

 

[Adeel Ahmad]

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

 

[Fightfish]

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

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

 

[Adeel Ahmad]

So that differentiates to 0 then

 

[Fightfish]

So that differentiates to 0 then

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

 

[Adeel Ahmad]

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

 

[Fightfish]

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

 

[Adeel Ahmad]

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

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?

 

[Fightfish]

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?

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##.

 

[Adeel Ahmad]

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?

 

[Adeel Ahmad]

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

 

[Fightfish]

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

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
[tex]f(x,y,z) = g(x,y,z) + h(x,y,z)[/tex]
Then, because partial derivation is linear, we have
[tex]f_{xyz} = g_{xyz} + h_{xyz}[/tex]
so we can find ##g_{xyz}## and ##h_{xyz}## separately and combine the results to get ##f_{xyz}##.

 

[Adeel Ahmad]

Ohh ok I got it now. Thank you so much for bearing with me.