Partial Derivatives: Solve Homework Quickly

Adeel Ahmad
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Homework Statement


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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?
 
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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.
 
So if I take the partial derivative wrt x and the result has no x term what does that mean?
 
Adeel Ahmad said:
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?
 
Fightfish said:
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
 
The partial derivative means to take all other variables as constants - what happens when you differentiate a constant?
 
Fightfish said:
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.
 
Adeel Ahmad said:
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:
\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##?
 
Isn't that what you would get if you differentiate with respect to x?
 
  • #10
Adeel Ahmad said:
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##.
 
  • #11
So that differentiates to 0 then
 
  • #12
Adeel Ahmad said:
So that differentiates to 0 then
Yup. Now look at the second term and see if you notice something similar.
 
  • #13
So for the second term, differentiating to z is 0 as well
 
  • #14
Yup. So immediately we conclude that ##f_{xyz}## is indeed ##0##.
 
  • #15
Fightfish said:
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?
 
  • #16
Adeel Ahmad said:
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##.
 
  • #17
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
Like, wouldn't I differentiate the entire function in terms of x?
 
  • #19
Adeel Ahmad said:
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
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
Ohh ok I got it now. Thank you so much for bearing with me.
 
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