Partial derivative of a single variable function

Click For Summary
SUMMARY

The discussion focuses on the application of the chain rule in calculating the partial derivative of a composite function, specifically U(x,y) = f(y + 2x). When taking the partial derivative with respect to x, the expression transforms into a total derivative due to the dependency of y + 2x on x. The correct formulation is given by the chain rule, resulting in ∂U/∂x = (∂f/∂g) * (∂g/∂x), where g(x,y) = y + 2x. This highlights the necessity of recognizing y + 2x as a function of both variables x and y.

PREREQUISITES
  • Understanding of partial derivatives and their notation
  • Familiarity with the chain rule in calculus
  • Basic knowledge of composite functions
  • Ability to differentiate functions of multiple variables
NEXT STEPS
  • Study the chain rule in depth, focusing on its application to composite functions
  • Practice calculating partial derivatives using examples of multiple variable functions
  • Explore the concept of total derivatives and their differences from partial derivatives
  • Learn about the implications of function composition in multivariable calculus
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and multivariable functions, as well as educators looking for clear explanations of the chain rule and partial derivatives.

wumple
Messages
57
Reaction score
0
So I don't understand why if you have something like U(x,y) = f(y+2x)

and you take [tex]\frac{\partial U}{\partial x}<br /> = \frac{\partial f}{\partial x}[/tex]

you get [tex]\frac{df}{d(y+2x)} * \frac{d(y+2x)}{dx}[/tex]

Why does the partial derivative just change to the total derivative for one variable? It seems like you just treat x and y+2x as the same variable? thanks for any help!
 
Physics news on Phys.org
Define a function g by g(x,y)=y+2x for all x,y. You have specified that the relationship between U, f and g is given by
$$U(x,y)=f(g(x,y))=(f\circ g)(x,y)$$ for all x,y. So ##U=f\circ g##, and therefore $$\frac{\partial U}{\partial x}=\frac{\partial (f\circ g)}{\partial x}.$$ Since we're dealing with a composition of functions, we need the chain rule. It tells us that the right-hand side is $$=f'\, \frac{\partial g}{\partial x}.$$ This is often written as $$\frac{\partial f}{\partial g}\frac{\partial g}{\partial x}.$$ The g in the denominator can serve as a reminder that f' is to be evaluated at g(x,y).
 
Last edited:
Thanks! I just wasn't thinking of y+2x as a separate function of x and y.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K