I have seen some statements and proofs of multivariable chain rule in various sites. I "somewhat" grasp them but seems too complicated for me to fully understand them.

To make my life easy, I have come up with a simple statement and a simple "rigorous" proof of multivariable chain rule. Please explain to what extent it is plausible.

PLEASE NOTE: In my statement of multivariable chain rule "$f[x(t),y(t)]$ is differentiable at $t=a$" is a condition rather than a provable result. I think it is the only way in which my statement differs from the usual statement.

I am a graduate Physics student and everywhere in my text (Electricity and Magnetism, Thermodynamics, etc) there is no mention of differentiability even though multivariable chain rule is used quite often. It seems to me the book just assumes that all functions used in the book are differentiable everywhere.

So with this little change in the statement, I do not think it will have any affect on my rigorous Physics study. Am I right?


Statement: If $f[x(t),y(t)]$, $x(t)$ and $y(t)$ are differentiable at $t=a$; and

$f(x,y)$ is differentiable at $x(t)=x(a)$ and $y(t)=y(a)$;

then at $t=a$

$$\dfrac{df[x(t),y(t)]}{dt}=\dfrac{\partial f[x(t),y(t)]}{\partial x(t)}\ \dfrac{dx(t)}{dt}+\dfrac{\partial f[x(t),y(t)]}{\partial y(t)}\ \dfrac{dy(t)}{dt}$$



\begin{align} \Delta f[x,y]&=f[x+\Delta x, y+\Delta y]-f[x,y]\\ &=f[x+\Delta x, y+\Delta y]-f[x,y+\Delta y]+f[x,y+\Delta y]-f[x,y]\\ &=\delta f_x[x,y]+\delta f_y[x,y]\\ \Rightarrow\ \Delta f[x(t),y(t)]&=\delta f_x[x(t),y(t)]+\delta f_y[x(t),y(t)]\\ \Rightarrow \dfrac{\Delta f[x(t),y(t)]}{\Delta t}&=\dfrac{\delta f_x[x(t),y(t)]}{\delta x(t)}\dfrac{\Delta x(t)}{\Delta t}+...\\ \Rightarrow \lim\limits_{\Delta t \to 0} \dfrac{\Delta f[x(t),y(t)]}{\Delta t}&= \lim\limits_{\Delta t \to 0} \left( \dfrac{\delta f_x[x(t),y(t)]}{\delta x(t)}\dfrac{\Delta x(t)}{\Delta t} \right)+...\\ \Rightarrow \lim\limits_{\Delta t \to 0} \dfrac{\Delta f[x(t),y(t)]}{\Delta t}&= \lim\limits_{\Delta t \to 0} \left( \dfrac{\delta f_x[x(t),y(t)]}{\delta x(t)} \right) \lim\limits_{\Delta t \to 0} \left( \dfrac{\Delta x(t)}{\Delta t} \right)+...\\ &\text{}\\ &\text{It is given that $x(t)$ is differentiable at $t=a$.}\\ &\text{Therefore $\lim\limits_{\Delta t \to 0} \dfrac{\Delta x(t)}{\Delta t}$ exists.}\\ &\text{Therefore when $\Delta t \to 0$, $\Delta x(t) \to 0$.}\\ &\text{}\\ \Rightarrow \lim\limits_{\Delta t \to 0} \dfrac{\Delta f[x(t),y(t)]}{\Delta t}&= \lim\limits_{\Delta x(t) \to 0} \left( \dfrac{\delta f_x[x(t),y(t)]}{\delta x(t)} \right) \lim\limits_{\Delta t \to 0} \left( \dfrac{\Delta x(t)}{\Delta t} \right)+...\\ &\text{}\\ &\text{It is given that $f[x(t),y(t)]$, $x(t)$ and $y(t)$ are differentiable at $t=a$;} \\ &\text{and $f(x,y)$ is differentiable at $x(t)=x(a)$ and $y(t)=y(a)$}\\ &\text{}\\ &\text{Therefore we can replace the limits with derivatives.}\\ &\text{}\\ \Rightarrow \dfrac{df[x(t),y(t)]}{dt} &= \dfrac{\partial f_x[x(t),y(t)]}{\partial x(t)}\ \dfrac{dx(t)}{dt} +...\\ \end{align}

So this is the statement and proof I have come up with. Again, please explain to what extent is it plausible (whether it is completely or partially rigour).

  • 2
    $\begingroup$ $(+1)$ for the amazing coding, considering you are relatively new to this site! :D $\endgroup$
    – Mr Pie
    Commented Feb 26, 2019 at 11:16
  • 1
    $\begingroup$ You can easily make up an example where the partial derivatives exist but the function is not differentiable. Essentially the reason is that those two directions $x$ and $y$ are arbitrary. It can fail to be differentiable in some other direction. $\endgroup$ Commented Feb 26, 2019 at 11:29
  • $\begingroup$ Thank you for pointing out one limitation. It seems to me that I need to listen to a lecture on differentiability of multivariable functions. I need to replace the statement "[ ] exists at $t=a$" with "$f(x,y)$ is differentiable at $x(t)=x(a)$ and $y(t)=y(a)$". If I do that, is everything else fine? $\endgroup$
    – Joe
    Commented Feb 26, 2019 at 11:46
  • 1
    $\begingroup$ I'll let someone else comment on that. While I likely could go through it, I haven't touched multivariable calculus in years (my specialty is abstract algebra) so I might miss something. $\endgroup$ Commented Feb 26, 2019 at 11:48

2 Answers 2


This is not rigorous at all. For one thing, you have not even defined most of your notation: what do $\Delta x(t)$, $\delta f_x(x,y)$, and so on mean? Even filling in reasonable guesses for what the notation means, there are serious issues. For instance, if $x(t)$ is a constant function, then it would seem that what you are referring to as $\delta x(t)$ is always $0$, so you cannot divide by it. There is also an issue that the difference $f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)$ is taken at $y+\Delta y$ instead of at $y$, and so you cannot expect it to be well-approximated using a partial derivative of $f$ at $(x,y)$ unless you know that partial derivative is continuous.

At best, what you have written is a sketch of a proof of the chain rule under significantly stronger hypotheses than you have stated.


At the moment your proof is over-complicated and you have not defined the meaning of many of your operators. You need to be careful to draw a distinction between when you are defining the meaning of an operation (which you should state as a definition) and when you are using rules of algebra to say something about that operation. There are some other problems (pointed out in detail by other commentators), and these mistakes probably stem from the fact that your proof is still much more complicated than it needs to be.

Here is an example of a simple proof structure for the multivariate chain rule, for a multivariate function of arbitrary dimension. This one is not a "rigorous" proof, since I have not gone to the effort of tightening up the cases where the denominators in the expressions are zero (which are trivial cases anyway). Nevertheless, if you were to tighten up these conditions then something like this method should allow you to construct a proof of the result. As you can see, all that is really happening is that you are expanding out the term $f(\mathbf{h}(t+\Delta))$ into a sum where you alter one argument value at a time.

THEOREM: Consider a multivariate function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and a vector $\mathbf{h} = (h_1,...,h_n)$ composed of univariate functions $h_i: \mathbb{R} \rightarrow \mathbb{R}$. We define $g: \mathbb{R} \rightarrow \mathbb{R}$ to be the composition of these functions, given by: $$g(t) = f(\mathbf{h}(t)) = f(h_1(t),...,h_n(t)) \quad \quad \quad \text{for all } t \in \mathbb{R}.$$ If $f$ is differentiable at the point $\mathbf{h}(t)$ and $\mathbf{h}$ is differentiable at the point $t$ then we have: $$\frac{dg}{dt}(t) = \nabla f(\mathbf{h}(t)) \cdot \frac{d \mathbf{h}}{dt}(t).$$

PROOF: For all $t$ and $\Delta$ we will define the vector: $$\mathbf{h}_*^{(i)} = (h_1(t+\Delta),...,h_i(t+\Delta),h_{i+1}(t),...,h_n(t)),$$ where we add $\Delta$ to the argument value for the first $i$ elements. Using this notation we can write: $$f(\mathbf{h}(t + \Delta)) = f(\mathbf{h}(t)) + \sum_{i=1}^n \Big[ f(\mathbf{h}_*^{(i)}) - f(\mathbf{h}_*^{(i-1)}) \Big].$$ Defining $\Delta_*^{(i)} \equiv h_{i}(t+\Delta) - h_i(t)$ we also have: $$f(\mathbf{h}_*^{(i)}) - f(\mathbf{h}_*^{(i-1)}) = f(\mathbf{h}_*^{(i-1)} + \Delta_*^{(i)} \mathbf{e}_i) - f(\mathbf{h}_*^{(i-1)}).$$ Now, using the definition of the derivative, and noting that $\Delta \rightarrow 0$ implies $\Delta_*^{(i)} \rightarrow 0$, we get: $$\begin{equation} \begin{aligned} \frac{d g}{d t} (\mathbf{x}) &= \lim_{\Delta \rightarrow 0} \frac{g(t + \Delta) - g(t)}{\Delta} \\[6pt] &= \lim_{\Delta \rightarrow 0} \frac{f(\mathbf{h}(t + \Delta)) - f(\mathbf{h}(t))}{\Delta} \\[6pt] &= \lim_{\Delta \rightarrow 0} \sum_{i=1}^n \frac{f(\mathbf{h}_*^{(i)}) - f(\mathbf{h}_*^{(i-1)})}{\Delta} \\[6pt] &= \lim_{\Delta \rightarrow 0} \sum_{i=1}^n \frac{f(\mathbf{h}_*^{(i)}) - f(\mathbf{h}_*^{(i-1)})}{h_{i}(t+\Delta) - h_i(t)} \cdot \frac{h_{i}(t+\Delta) - h_i(t)}{\Delta} \\[6pt] &= \sum_{i=1}^n \Bigg( \lim_{\Delta\rightarrow 0} \frac{f(\mathbf{h}_*^{(i)}) - f(\mathbf{h}_*^{(i-1)})}{h_{i}(t+\Delta) - h_i(t)} \Bigg) \cdot \Bigg( \lim_{\Delta \rightarrow 0} \frac{h_{i}(t+\Delta) - h_i(t)}{\Delta} \Bigg) \\[6pt] &= \sum_{i=1}^n \Bigg( \lim_{\Delta \rightarrow 0} \frac{\partial f}{\partial h_i}(\mathbf{h}_*^{(i)}) \Bigg) \cdot \frac{d h_i}{dt}(t) \\[6pt] &= \sum_{i=1}^n \frac{\partial f}{\partial h_i}(\mathbf{h}(t)) \cdot \frac{d h_i}{dt}(t) \\[6pt] &= \nabla f(\mathbf{h}(t)) \cdot \frac{d \mathbf{h}}{dt}(t). \\[6pt] \end{aligned} \end{equation}$$

To be more rigorous, we note that there may be cases where one or more of the term $\Delta_*^{(i)}$ are zero, in which case we cannot divide through by these terms in the denominator. This does not cause problems because the term in the summation is zero in this case, so the whole term can be removed. This establishes the desired result. $\blacksquare$

  • $\begingroup$ Your proof is still badly wrong, due to the second issue I mentioned. $\lim_{\Delta_*^{(i)} \rightarrow 0} \frac{f(\mathbf{h}_*^{(i-1)} + \Delta_*^{(i)} \mathbf{e}_i) - f(\mathbf{h}_*^{(i-1)})}{\Delta_*^{(i)}}$ is $\frac{\partial f}{\partial h_i}(\mathbf{h}_*^{(i-1)})$, not $\frac{\partial f}{\partial h_i}(\mathbf{h}(t))$. You need to use the fact that $f$ is differentiable, not just that it has partial derivatives. $\endgroup$ Commented Mar 1, 2019 at 16:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .