Finding the Second Partial Derivative of a Multivariable Function

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SUMMARY

The discussion centers on verifying that the function z(x,y) = cos(xy) satisfies the equation (∂z/∂x)y + (∂z/∂y)x = (x+y)((∂²z/∂x∂y) + xyz). Participants clarify that demonstrating z(x,y) as a solution involves substituting it into both sides of the equation and confirming their equality. The first derivatives are calculated as ∂z/∂x = -ysin(xy) and ∂z/∂y = -xsin(xy), leading to the need for the second mixed partial derivative ∂²z/∂x∂y.

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  • Understanding of multivariable calculus concepts, particularly partial derivatives.
  • Familiarity with trigonometric functions and their derivatives.
  • Knowledge of the chain rule in differentiation.
  • Ability to manipulate and simplify algebraic expressions involving derivatives.
NEXT STEPS
  • Calculate the second mixed partial derivative ∂²z/∂x∂y for z(x,y) = cos(xy).
  • Review the properties of partial derivatives in multivariable functions.
  • Study the application of the chain rule in the context of multivariable calculus.
  • Explore examples of verifying solutions to differential equations involving multivariable functions.
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Students and educators in mathematics, particularly those studying multivariable calculus, as well as anyone interested in understanding the verification of solutions to differential equations.

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Homework Statement



Show z(x,y) = cos(xy) is a solution of

(∂z/∂x)y + (∂z/dy)x = (x+y) ( (∂2z/∂x∂y) + xyz)

(question also attached if it makes it clearer)

The Attempt at a Solution



∂z= (∂z/∂x)ydx + (∂z/dy)xdy

∂z/∂x = -ysin(xy)
∂z/∂y = -xsin(xy)

what does it mean show it is a solution? any tips appreciated
 

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chemphys1 said:
what does it mean show it is a solution? any tips appreciated
It means that choice of z(x,y) satisfies the equation. I.e plug in z(x,y) into LHS and into the RHS and they should be equal.
 
So for ##z(x,y) = \cos(xy) ##, what is ##\frac{∂^2z}{∂x∂y}##?
 

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