Partial Derivatives: Solve Homework Quickly

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SUMMARY

The discussion focuses on solving partial derivatives in multivariable calculus, specifically addressing the confusion around differentiating functions with respect to multiple variables. The key takeaway is that the partial derivative is a linear operator, allowing the differentiation of individual terms separately. The participants clarify that if a term does not contain the variable being differentiated, its partial derivative will be zero. Ultimately, the conclusion is that for the function discussed, the partial derivative with respect to x, y, and z results in zero.

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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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