Implicit 'higher' differentiation

[City88]

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?

 

[Defennder]

What is the implicit function you are given?

 

[HallsofIvy]

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+ ye^z= 1, then the partial derivative, with respect to x, would be given by 3z+ 3xz_x+ ye^zz_x= 0.

Differentiating that a second time, with respect to x, 3z<sub>x+ 3zx+ 3xzxx+ ye^z(zx)²+ ye^zzxx= 0.

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