What is the Technique for Solving Partial Differentiation in Calculus 2?

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SUMMARY

The discussion focuses on solving a problem related to partial differentiation in a Calculus 2 course. The equation provided, (x^2+y^2+z^2)^-1/2=V, requires proving that the sum of the second partial derivatives, dv^2/dx^2 + dv^2/dy^2 + dv^2/dz^2, equals zero. Participants clarify that the problem involves taking ordinary derivatives while treating the other variables as constants, rather than dealing with differential equations.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with ordinary derivatives
  • Basic knowledge of multivariable calculus
  • Concepts from Calculus 2 curriculum
NEXT STEPS
  • Practice solving problems involving partial derivatives
  • Review the chain rule for multivariable functions
  • Study the application of second derivatives in multivariable calculus
  • Explore the relationship between partial derivatives and gradient vectors
USEFUL FOR

Students enrolled in Calculus 2, educators teaching multivariable calculus, and anyone seeking to strengthen their understanding of partial differentiation techniques.

Isma
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i ve never read partial DE...nd i don't kno how to do this question i got in homework...pleasez help
(x^2+y^2+z^2)^-1/2=V
prove dv^2/dx^2 + dv^2/dy^2 + dv^2/dz^2 = 0
(i wrote "d" for partial differential)
i know its a basic question but i can't understand the technique
 
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What course is this for?
It doesn't matter that you have "never read partial DE"- this problem has nothing to do with differential equations. It has to do with taking the derivative. If you mean you have not done partial derivatives, the derivative of V with respect to x is just the ordinary derivative, treating y and z as constants. Similarly, the derivative of V with respect to y is just the ordinary derivative, treating x and z as constants; the derivative of V with respect to z is just the ordinary derivative, treating x and y as constants.

Find the second derivative of V with respect to each variable, and add them!
 
thx...that was easy
it is for Calculus 2 course
 

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