# Implicit differentiation

• Oct 23rd 2009, 01:36 PM
kodos
Implicit differentiation
Let F: $R^2 \rightarrow R$ be a function $C^2$ for which F(x+y,y+z)=0 defines implicitly z=f(x,y) around a point (a,b,c). Verify that in (a,b)
dz/dx-dz/dy=1

I'm having difficulty with getting rid of some derivatives, they should cancel out I guess.

(It's a bit heavy to read but please take a look at my work)

First, I apply the chain rule with u=g(x,y)=x+y, v=h(x,y)

and get the expressions for dz/dx and dz/dy

dF/dx=dF/du dg/dx+dF/dv dh/dx=dF/du+dF/dv dz/dx

solve for dz/dx,

dz/dx=(dF/dx-dF/du)/(dF/dv)

Then,

dF/dy=dF/du+dF/du dz/dy

dz/dy={dF/dy-dF/du}/{dF/dv}

So dz/dx-dz/dy={dF/dx-dF/dy-2dF/du}/{dF/dv}=???

From dF/dz=dF/dv but dF/dz is no good.

How can I get rid of those derivatives?
• Oct 23rd 2009, 02:00 PM
TheEmptySet
Quote:

Originally Posted by kodos
Let F: $R^2 \rightarrow R$ be a function $C^2$ for which F(x+y,y+z)=0 defines implicitly z=f(x,y) around a point (a,b,c). Verify that in (a,b)
dz/dx-dz/dy=1

I'm having difficulty with getting rid of some derivatives, they should cancel out I guess.

(It's a bit heavy to read but please take a look at my work)

First, I apply the chain rule with u=g(x,y)=x+y, v=h(x,y)

and get the expressions for dz/dx and dz/dy

dF/dx=dF/du dg/dx+dF/dv dh/dx=dF/du+dF/dv dz/dx

solve for dz/dx,

dz/dx=(dF/dx-dF/du)/(dF/dv)

Then,

dF/dy=dF/du+dF/du dz/dy

dz/dy={dF/dy-dF/du}/{dF/dv}

So dz/dx-dz/dy={dF/dx-dF/dy-2dF/du}/{dF/dv}=???

From dF/dz=dF/dv but dF/dz is no good.

How can I get rid of those derivatives?

You have an error in your $\frac{dF}{dy}$

You should get

$\frac{dF}{dy}=\frac{dF}{du}\frac{du}{dy}+\frac{dF} {dv}\frac{dv}{dy}+\frac{dF}{dv}\frac{dv}{dz}\frac{ dz}{dy}=0$