# Thread: Backwards Chain Rule

1. ## Backwards Chain Rule

I am too lazy to search for counter examples if somebody can find one that would be excellent. More excellent if he can proof it.

Let f be a differenciable function at c and let g be a function which is defined on an open interval containing f(c) such that (g o f) is differenciable at c. Show that g is differenciable at c.

2. ## Re:

RE:

The Backwards Chain Rule...
18.3 Tricks of Integration

3. Example 1:
Let g(x) = 1/x
Let f(x) = x + 1
Let c = 0

Example 2:
Let g(x) = |x - 3|
Let f(x) = 2x
Let c = 3

Argument:
If f(x) is some function defined and differentiable on a closed interval [a,b] containing the point x = c, where f(c) = d != c and f'(c) = e, and if g(x) is some function defined on a piecewise, differentiable domain D = {x | x !=c} containing the point x = d, where g'(d) = n, then it holds that
Although
d/dx(g o f)(c) = g'(f(c))*f'(c) = g'(d)*e = ne <-- is defined
d/dx(g(c)) <-- is undefined

I'm not sure that counts as a formal proof. I've learned proofs on my own (much of it on this site), so I cannot guarantee that there are no flaws in my logic.

4. Originally Posted by qbkr21
RE:

The Backwards Chain Rule...
18.3 Tricks of Integration
My problem is more complicated then that. I cannot just use the rule without showing why it will work.

5. Consider f(x)=x^2, g(x)=|x| and c=0.
Both fog and gof are differentiable at c.

6. Those examples certainly disprove it.