Implicit 'higher' differentiation

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Hi,
I'm working on a cal III problem involving implicit differentiation.
I have to find the second order partial derivative of an implicit function, basically:
[tex]\partial[/tex]2f
[tex]\partial[/tex]x2

now, I know that for a single order [tex]\partial[/tex]f/[tex]\partial[/tex]x, I would simply use the chain rule property:
[tex]\partial[/tex]f = -[tex]\partial[/tex]F/[tex]\partial[/tex]x
[tex]\partial[/tex]x ... [tex]\partial[/tex]F/[tex]\partial[/tex]f

But now, how would I find
[tex]\partial[/tex]2f
[tex]\partial[/tex]x2
for an implicit equation?
 
Last edited:

Answers and Replies

  • #2
Defennder
Homework Helper
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What is the implicit function you are given?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Hi,
I'm working on a cal III problem involving implicit differentiation.
I have to find the second order partial derivative of an implicit function, basically:
[tex]\partial[/tex]2f
[tex]\partial[/tex]x2

now, I know that for a single order [tex]\partial[/tex]f/[tex]\partial[/tex]x, I would simply use the chain rule property:
[tex]\partial[/tex]f = -[tex]\partial[/tex]F/[tex]\partial[/tex]x
[tex]\partial[/tex]x ... [tex]\partial[/tex]F/[tex]\partial[/tex]f

But now, how would I find
[tex]\partial[/tex]2f
[tex]\partial[/tex]x2
for an implicit equation?

Do the same thing again. For example, if the function were z, given by 3xz+ yez= 1, then the partial derivative, with respect to x, would be given by 3z+ 3xzx+ yezzx= 0.

Differentiating that a second time, with respect to x, 3zx+ 3zx+ 3xzxx+ yez(zx)2+ yezzxx= 0.

You can solve that for zxx in terms of x, y, z, and zx.
 

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