- #1
bobsmith76
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Homework Statement
I don't understand why
∂f/∂x = xy = y
whereas
∂f/∂x = x2 + y2 = 2x
Why does the y disappear in the second but not in the first?
Confusion with Partial Derivatives: Why does y disappear? | Explained
- Thread starter bobsmith76
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SUMMARY
The discussion clarifies the concept of partial derivatives, specifically addressing why the variable y appears in the derivative of the function f(x, y) = xy but not in f(x, y) = x² + y². When differentiating ∂f/∂x = xy, y is treated as a constant, resulting in ∂/∂x (xy) = y. Conversely, in ∂f/∂x = x² + y², the term y² is independent of x, leading to ∂/∂x (x² + y²) = 2x, where the derivative of y² is zero.
PREREQUISITES- Understanding of basic calculus concepts, particularly differentiation.
- Familiarity with the notation of partial derivatives, specifically ∂/∂x.
- Knowledge of treating variables as constants during differentiation.
- Basic algebra skills for manipulating expressions involving multiple variables.
- Study the rules of differentiation, focusing on partial derivatives in multivariable calculus.
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- Explore examples of functions with multiple variables to practice calculating partial derivatives.
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Students of calculus, educators teaching multivariable calculus, and anyone seeking to deepen their understanding of partial derivatives and their applications.
I don't understand why
∂f/∂x = xy = y
whereas
∂f/∂x = x2 + y2 = 2x
Why does the y disappear in the second but not in the first?
- #2
tiny-tim
Science Advisor
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hi bobsmith76! 
you mean? …
in each case, y is treated as a constant
in the first, it's multiplied, so it stays; in the second, it's on its own, so its derivative is zero
you mean? …
∂/∂x (xy) = y
∂/∂x (x2 + y2) = 2x
∂/∂x (x2 + y2) = 2x
in each case, y is treated as a constant
in the first, it's multiplied, so it stays; in the second, it's on its own, so its derivative is zero
∂/∂x means differentiating wrt x while keeping all other variables constant
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