# Implicit 'higher' differentiation

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:
$$\partial$$2f
$$\partial$$x2

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

But now, how would I find
$$\partial$$2f
$$\partial$$x2
for an implicit equation?

Last edited:

Defennder
Homework Helper
What is the implicit function you are given?

HallsofIvy
Homework Helper
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:
$$\partial$$2f
$$\partial$$x2

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

But now, how would I find
$$\partial$$2f
$$\partial$$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.