Do these two partial derivatives equal each other?

Click For Summary
SUMMARY

The discussion centers on the equality of partial derivatives in the context of the function f(x, y, z). Specifically, it addresses whether the partial derivative of f with respect to z, while holding x constant, is equal to the partial derivative of f with respect to z, while holding y constant. The consensus is that these derivatives are indeed equal, as the notation indicating which variable is held constant is unnecessary when differentiating with respect to z. The conversation also highlights the importance of context in defining functions and their derivatives, particularly in complex scenarios involving multiple variables.

PREREQUISITES
  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with differential notation and the chain rule
  • Knowledge of implicit function definitions in calculus
  • Basic concepts of thermodynamics and their mathematical representations
NEXT STEPS
  • Study the properties of partial derivatives in multivariable calculus
  • Learn about implicit differentiation and its applications
  • Explore the chain rule for functions of multiple variables
  • Investigate the use of partial derivatives in thermodynamics and physics
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who require a solid understanding of multivariable calculus, particularly in the context of partial derivatives and their applications in various fields.

thegirl
Messages
40
Reaction score
1
take the function f(x,y,z)

s.t dF=(d'f/d'x)dx+(d'f/d'y)dy+(d'f/d'z)dz=0 where "d'" denotes a curly derivative arrow to show partial derivatives
Mod note: Rewrote the equation above using LaTeX.
$$df = (\frac{\partial f}{\partial x} ) dx + (\frac{\partial f}{\partial y} ) dy + (\frac{\partial f}{\partial z} ) dz = 0$$

Is this statement true? (d'f/d'z)x=(d'f/d'z)y (the partial derivative of the function with respect to z at a constant x equal to the partial derivative of the function with respect to z with y as a constant?

Could someone explain this to me? Are those partial derivatives equal to each other?
 
Last edited by a moderator:
Physics news on Phys.org
thegirl said:
take the function f(x,y,z)

s.t dF=(d'f/d'x)dx+(d'f/d'y)dy+(d'f/d'z)dz=0 where "d'" denotes a curly derivative arrow to show partial derivatives
Mod note: Rewrote the equation above using LaTeX.
$$df = (\frac{\partial f}{\partial x} ) dx + (\frac{\partial f}{\partial y} ) dy + (\frac{\partial f}{\partial z} ) dz = 0$$
Note that I changed dF as you wrote it to df, assuming you're talking about the function f.
thegirl said:
Is this statement true? (d'f/d'z)x=(d'f/d'z)y (the partial derivative of the function with respect to z at a constant x equal to the partial derivative of the function with respect to z with y as a constant?
Since df = 0, the f(x, y, z) = C, for some constant C.
##\frac{\partial f}{\partial z}## is the partial derivative of f with respect to z. It's redundant to say the partial of f with respect to z, with x held constant or with y held constant. Since you're differentiating with respect to z, x and y are held constant anyway.
thegirl said:
Could someone explain this to me? Are those partial derivatives equal to each other?
 
The notation that specifies which variables are held constant is only needed when it wouldn't be clear otherwise what function we're differentiating. For example, suppose that the 4 variables w,x,y,z satisfy the equations ##x+y+z=1## and ##w=2x+y^2-z##. If you write ##\frac{\partial w}{\partial x}##, it's not entirely clear if the function you want to differentiate is the g defined by ##g(x,y)=2x+y^2-(1-y-x)## or the h defined by ##h(x,z)=2x+(1-x-z)^2-z##. If it's the former, you can write ##D_1g(x,y)## or ##\left(\frac{\partial w}{\partial x}\right)_y##. If it's the latter, you can write ##D_1h(x,z)## or ##\left(\frac{\partial w}{\partial x}\right)_z##.

The given equations define at least two functions (h and g) implicitly. The need to write out "what variable is held constant" arises when you chose not to introduce new symbols (like h and g) for those functions, and instead just use the symbol w for one of them.
 
You title this "Do these two partial derivatives equal each other" but what you have are three partial derivatives and what you are equating are differentials, not derivatives. Just as in single variable Calculus, we start by defining the "derivative" dy/dx= f'(x) but then define the "differentials" by "dy= f'(x)dx", so we define the "differential" for a function of several variables.

We can do this: Let f(x,y,z) be a function of the three variables x, y, and z, and suppose that each of x, y, and z is a function of some other variable, t. Then f can be thought of as a function of the single variable t and, by the chain rule, \frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}. Now we can write that in "differential form" as df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz
 
HallsofIvy said:
You title this "Do these two partial derivatives equal each other" but what you have are three partial derivatives and what you are equating are differentials, not derivatives.
I think she was trying to ask if the partial derivative of f "with respect to z, and with x constant" is equal to the partial derivative of f "with respect to z, and with y constant". Of course, there's no need to say that one of the variables are held constant if f can only denote one function. The specification of what variable is held constant is meant to be the last piece of information you need to know what function f is. (See my previous post for some details).

This concept is often used in physics books (especially in thermodynamics), where it's convenient to have the same symbol for several different functions whose values are meant to be interpreted as values of the same physical quantity.
 

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K