Exact Differential: Show f(z)dz is Exact

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SUMMARY

The discussion confirms that the differential form f(z)dz is exact if and only if the function f(z) possesses a primitive. This concept is rooted in multivariable calculus, specifically involving functions of two variables, x and y. An expression A(x,y)dx + B(x,y)dy is classified as an exact differential when there exists a differentiable function F(x,y) such that dF = A(x,y)dx + B(x,y)dy. The necessary condition for exactness is that the partial derivatives satisfy the equality ∂A/∂y = ∂B/∂x.

PREREQUISITES
  • Understanding of multivariable calculus
  • Familiarity with differential forms
  • Knowledge of partial derivatives
  • Concept of primitives in calculus
NEXT STEPS
  • Study the properties of exact differentials in multivariable calculus
  • Learn about the existence of primitives for functions of multiple variables
  • Explore the implications of the equality ∂A/∂y = ∂B/∂x in differential forms
  • Investigate applications of exact differentials in physics and engineering
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Students and professionals in mathematics, particularly those focusing on multivariable calculus, as well as physicists and engineers applying these concepts in practical scenarios.

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Q. Show that f(z)dz defined in a region is exact if and only if f(z) has a primitive.
 
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Re: exact differential

ssh said:
Q. Show that f(z)dz defined in a region is exact if and only if f(z) has a primitive.

Usually thye concept of 'exact differential' refers to a multivariable function. In case of two variables x and y, an expression like... $\displaystyle A(x.y)\ dx + B(x,y)\ dy\ (1)$ ... where A(*,*) and B(*,*) are defined in a field D, is called exact differential if it exist an F(x,y) differentiable in D for which is... $\displaystyle dF = A(x,y)\ dx + B(x,y)\ dy\ (2)$

The expression (1) is an exact differential if and only if $A(x,y)$, $B(x,y)$, $\displaystyle \frac{\partial A}{\partial y}$ and $\displaystyle \frac{\partial B}{\partial x}$ are continuos and is... $\displaystyle \frac{\partial A}{\partial y}= \frac{\partial B}{\partial x}\ (3)$Kind regards $\chi$ $\sigma$
 

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