# What's wrong with this proof of the chain rule

1. May 26, 2008

### quasar987

Can anyone see where the flaw is in the developement below, where I prove that (g o f)'(x)=g'(f(x)) instead of g'(f(x))f'(x), as it should be.

Consider the usual hypothese under which the chain rule for real-valued function applies.

Consider $\epsilon>0$. Since g is differentiable at f(x_0), there exists $\delta_1(\epsilon)>0$ such that $|y-f(x_0)|<\delta_1(\epsilon)$ implies $|g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq\epsilon|y-f(x_0)|$

On the other hand, f being differentiable at x_0, enjoys the Lipschitz property there. That is to say, there exists positive constants c and M such that $|x-x_0|<c$ implies $|f(x)-f(x_0)|\leq M|x-x_0|$.

So for $\delta=\min(c,\delta_1(\epsilon/M)/M)$, we have that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|\leq M|x-x_0|<M\delta\leq M\delta_1(\epsilon/M)/M= \delta_1(\epsilon/M)$

which in turns implies that

$$|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq(\epsilon/M) |f(x)-f(x_0)|\leq (\epsilon/M)M|x-x_0| = \epsilon |x-x_0|$$

And this is equivalent to saying that (g o f)'(x) = g'(f(x))...

Last edited: May 26, 2008
2. May 26, 2008

### quasar987

Ok, I see it.