# Derivative of a signal $y(t)$ wrt to another signal $x(t)$

I am running a sensitivity study on the model $$y(t) = x(t - \tau)$$ where $$y(t)$$ and $$x(t)$$ are 2 time signals and $$\tau$$ a time lag. Basically I want to study the sensitivity of $$y$$ to a change in $$x$$. Note that the model is very simple just for the sake of illustrating my question.

What I think about is the derivative of $$y$$ with respect to $$x$$, but I cannot see clearly how this could be done. I know the result is 1, but the way I see it is:

$$$$\frac{\partial y}{\partial x(t-\tau)}(t) = \frac{\partial y}{\partial t}(t)\frac{\partial t}{\partial x(t - \tau)} = \frac{\partial x}{\partial t}(t-\tau) \frac{\partial t}{\partial x(t-\tau)} = 1$$$$ so that a change in $$x$$ at $$t-\tau$$ is seen by the same change at a future time $$t$$ in $$y$$.

Is that a rigorous and mathematical way to look at it? I know that I might have messed up the equation and the time arguments, but I would like to learn the rigorous way to write such a problems.

EDIT:

Maybe to show the general case, I am interested in the following analytical models: $$y(t) = \sum_i f_i^{p_i}(x(t - \tau_i))$$ where $$f_i$$ is an elementary function. For instance if $$f_i(x) = x \ \forall i$$, then $$y(t) = \sum_i x^{p_i}(t - \tau_i)$$. What i am interested in is sensitivities of $$y$$ wrt to the different $$x_i$$'s.

• Is there some connection (for example a model) of how $y$ and $x$ are related? I'm curious, for example, if you changed $x$ just a bit, how that would propagate to $y$. Are these measured data? Sep 13 at 17:03
• If, on the other hand, all you want to do is explore how $y$ is related to a lagged $x$, you might explore phase relationships in the frequency domain. Sep 13 at 18:26
• @Aruralreader the model is that $y(t) = x(t-\tau)$ ... Sep 13 at 19:50
• So, for two given time series x(t) and y(t) we want to verify the model $y(t)=x(t-\tau)$, where $\tau$ is a given (?) time lag. If the model was perfectly well describing the data then plotting y vs. x data points we'd see they all are on a straight line (y=x). If the model is not perfect then the data points would be forming some kind of cloud, but one can find a fit using least squares fitting, so it will be y = a x + b + $\sigma$, where $\sigma$ is the average error of the fit. These three parameters: a, b, sigma is what describes the sensitivity of y to x, on average, for given data set. Sep 14 at 0:49
• If one signal is a time shift of the other, perhaps with an amplitude scaling, this would show up quickly in a (windowed) cross-correlation. Windows and the frequency domain will be your friends. See any signal processing book to get started. Sep 14 at 1:10

In essence, what you are asking is the ratio of how $$y$$ changes at $$t$$ to how $$x$$ changes at $$t$$. That is: $$\frac{dy}{dx}(t) = \frac{\frac{dy}{dt}(t)}{\frac{dx}{dt}(t)}.$$ Using that $$y(t)=x(t-\tau)$$, you then get that this is equal to $$\frac{dy}{dx}(t) = \frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)}.$$ But this is definitely not equal to one: For example, for $$x(t)=t^2$$, you would obtain $$\frac{dy}{dx}(t) = \frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)} = \frac{t-\tau}{t}.$$ It is true, however, that $$\lim_{\tau\rightarrow 0}\frac{dy}{dx}(t) = \lim_{\tau\rightarrow 0}\frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)} = 1.$$

• Thanks for the answer. What I dont see clear is the third equation in the sense that I expected the derivative to be 0, since a change in $x$ at $t$ is only seen in $y$ at $t + \tau$, but not at $t$. Sep 14 at 6:03
• Can you also please check the updated question Sep 14 at 6:14
• @outlaw When you say "I expected the derivative to be zero, since a change in $x$ at $t$ is only seen in $y$ at $t+\tau$ but not $t$", you seem to imply a causal relationship. But that is not the question. The way you formulate it is as a correlation: How are changes at times $t$ and $t+\tau$ correlated, and my derivation shows you how. You would be right, however, if you thought that $x(t)$ is a random function, in which values are completely uncorrelated. In that case, however, you would already have trouble defining time derivatives. Sep 14 at 16:39
• I see your point. Can you point out a reference for the definition you used: $\frac{dy}{dx}(t) = \frac{\frac{dy}{dt}(t)}{\frac{dx}{dt}(t)}$ ? I know one can say this is the chain rule, but I dont see the mathematical derivation of it obviously. Sep 15 at 10:50
• I don't have a reference, but for sufficiently smooth functions, this makes sense if you consider $dx$, $dy$, $dt$ to be finite changes before making the transition to infinitesimals. Sep 15 at 17:33

After more thinking, I am answering my own question. What I was looking for is the following:

Given for example $$y(t) = x(t) + z(t-\tau_1)k(t)$$, where $$x(t), y(t), z(t), k(t)$$ time signals, I can write $$y(t) = F(t, x(t), z(t-\tau_1), k(t))$$ for some function $$F$$. I want to study the following:

For a fixed time $$t$$ \begin{align} & \lim_{\epsilon \to 0} \frac{F(t, x(t) + \epsilon, z(t-\tau_1), k(t)) - F(t, x(t),z(t-\tau_1), k(t))}{\epsilon} = \\ &\lim_{\epsilon \to 0} \frac{x(t) + \epsilon + z(t-\tau_1)k(t) - x(t) - z(t-\tau_1)k(t)}{\epsilon} = \lim_{\epsilon\to 0} \frac{\epsilon}{\epsilon} = 1. \end{align} This quantity is what I refer to $$\frac{\partial y}{\partial x(t)}(t)$$. For example also, \begin{align} \frac{\partial y}{\partial z(t-\tau_1)}(t) := &\lim_{\epsilon \to 0} \frac{F(t, x(t), z(t-\tau_1) + \epsilon, k(t)) - F(t, x(t),z(t-\tau_1), k(t))}{\epsilon} \\ & \lim_{\epsilon \to 0} \frac{(z(t-\tau_1)+\epsilon)k(t) - z(t-\tau_1)k(t)}{\epsilon} = k(t) \end{align}

The partial derivatives notations that I am using might make sense mathematically, but in terms of limits, that is what I was looking for.