I found this proof of the chain rule, and it looks very thorough and legitimate. I have a few questions concerning it though.

Is this in fact a legitimate proof of the chain rule? As the author mentions at the end of the reading, a lot of proofs in calculus books do the chain rule wrong.

On page 2, the author concludes that $f(g(x) + [g'(x) + v]h) = f(g(x)) + [f'(g(x)) + w] \cdot [g'(x) + +v]h$. I have stared at this line for the longest time now and I am not understanding how the author gets this.

Is the flawed proof wrong since he is assuming $g(x + h) - g(x)$ is nonzero?

  • $\begingroup$ I've never seen that 'flawed proof' before, in particular I'm not aware of any text book presenting it. It is wrong since it does not show that the derivative exists, rather draws conclusions from the assumption it does (among other things, including the question you mentioned). On the other hand I think the proof which is given is a bit too complicated, I refused to check the details because of this. For a straighforward nononsense proof have a look at (e.g.) Rudin's principle of mathematical analysis, thm 5.5 $\endgroup$
    – Thomas
    Apr 27, 2014 at 19:02

1 Answer 1


The flawed proof is wrong because of the reason you have given.

The proof in the quoted source is not easily understandable, since the author introduces several "variables", whereby it remains unclear what exactly a "variable" is. The following excerpt from a previous answer of mine implements the proof idea of your source in a (hopefully) more clear way.

We shall make use of the following principle (which works both ways): When the function $g$ is differentiable at $x_0$ then the "trend function" $$m_{g,x_0}(x):=\cases{{g(x)-g(x_0)\over \mathstrut x-x_0} \quad&$(x\ne x_0)$ \cr \mathstrut g'(x_0)& $(x=x_0)$\cr}$$ is continuous at $x_0$, and one has $$g(x)-g(x_0)=m_{g,x_0}(x)\cdot(x-x_0)\qquad\forall x\ .$$ When in addition a function $f$ is given which is differentiable at $g(x_0)$ then one can apply said principle to $f$ as well and write $$\eqalign{f\bigl(g(x)\bigr)-f\bigl(g(x_0)\bigr)&= m_{f,g(x_0)}\bigl(g(x)\bigr)\cdot \bigl(g(x)-g(x_0)\bigr)\cr &=m_{f,g(x_0)}\bigl(g(x)\bigr)\cdot\ m_{g,x_0}(x)\cdot(x-x_0)\ .\cr}$$ Since $g$ is continuous at $x_0$ we can conclude that the first two factors on the right hand side make up a function which is continuous at $x_0$ and assumes the value $f'\bigl(g(x_0)\bigr)\>g'(x_0)$ there. The chain rule follows.

  • $\begingroup$ Thank you. This is the clearest proof of the chain rule I've ever found. :) $\endgroup$ May 6, 2019 at 22:25

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