# Realising I don't truly understand differentiation when not stated explicitly.

Probably much too late in my maths 'career' to be asking this but, I don't fully understand differentiation in multiple dimensions when functions aren't stated explicitly. I have got by up to now, but I think it's time to learn properly.

Some examples:

In my PDE's course we are studying the wave equation and for one question we define $$u(x,t) = v(\xi,\eta),\,\, \xi = x+ct,\,\, \eta = x-ct$$ where $$c\in\mathbb{R}$$. And the following line is $$\partial_{x}u = \partial_{\xi}v+\partial_{\eta}v$$. I understand where this comes looking at the formula for differentiation, but get confused with the dependancies, as it is almost never explicitly written.

Also considering functions from multiple variables to one. Say for example you have a differentiable function $$f$$ and you try to calculate $$\frac{d}{dx}f(ax+bt)\quad \text{and}\quad \frac{d}{dt}f(ax+bt)$$ where $$a,b \in\mathbb{R}$$ and $$x,t$$ are variables. I believe the answers respectively are $$af'(ax+bt)\quad \text{and}\quad bf'(ax+bt).$$ But, I have no idea where $$f'$$ comes from in this context. Does it mean $$f' = \frac{d}{d(ax+bt)}?$$ If so why? How can I find some intuition with calculating non-explicit forms as such?

• Maybe I'm missing something, but can you further explain what you mean when you say that you get confused with the dependencies? Maybe give a specific example? May 24, 2019 at 11:57
• @PauloMourão So for the example I gave $\partial_{x}u = \partial_{\xi}v+\partial_{\eta}v$. Is it really, $\partial_{x}u(x,t) = \partial_{\xi}v(\xi ,\eta)+\partial_{\eta}v(\xi ,\eta)$? May 24, 2019 at 12:09
• Yes, precisely. That gives you an equation that depends on $x$ and $t$ on the L.H.S and an equation that depends on $\xi$ and $\eta$ on the R.H.S. The claim is that these are the same, given the relations that you started with May 24, 2019 at 12:11

I often find it to be more clear to write inputs explicitly, and also to use the notation $$D_i f$$ for the $$i$$th partial derivative of a function $$f$$. The function $$u$$ is defined so that $$u(x,t) = v(x + ct, x - ct) = v(\xi(x,t), \eta(x,t)),$$ where $$\xi(x,t) = x + ct$$ and $$\eta(x,t) = x - ct$$. From the multivariable chain rule, the first partial derivative of $$u$$ is \begin{align} D_1 u(x,t) &= D_1 v(\xi(x,t), \eta(x,t)) D_1 \xi(x,t) + D_2 v(\xi(x,t), \eta(x,t)) D_1 \eta(x,t)\\ &= D_1 v(x + ct, x - ct) \cdot 1 + D_2 v(x + ct, x - ct) \cdot 1. \end{align} This result could also be written as $$\partial_x u(x,t) = \partial_\xi v(x+ct, x - ct) + \partial_\eta v(x + ct, x- ct).$$

Regarding the second example, define $$F(x,t) = f(ax + bt) = f(h(x,t)),$$ where $$h(x,t) = ax + bt$$. From the chain rule, we have \begin{align} D_1 F(x,t) &= f'(h(x,t)) D_1 h(x,t) \\ &= f'(ax + bt) \cdot a. \end{align} This calculation could also be written as \begin{align} \partial_x F(x,t) &= f'(h(x,t)) \partial_x h(x,t) \\ &= f'(ax + bt) \cdot a. \end{align}

This can make more sense if you write it out with the chain rule. Let $$u=ax+bt$$. Then, with the chain rule, we get for $$\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}$$ with $$\frac{du}{dx}=a$$. So we get $$\frac{df}{dx}=a\frac{df}{du}$$ So your interpretation of $$f'=\frac{df}{d(ax+bt)}$$ is correct here.

• Note that the letter $f$ refers to two different functions in the first equation. I think it's more clear to give these functions different names. May 24, 2019 at 12:00
• @littleO This is my confusion, is the first $f$ really $f(ax+bt)$ and the second $f$ is $f(u)$? May 24, 2019 at 12:12

Maybe it is better if you use different letters for different functions... If fact you are defining $$F(x,t) = f(ax+bt)$$. The partial derivatives are related to the derivatives of $$F$$, while the total derivatives are related to $$f$$. For instance, as you say, $$\frac{\partial F}{\partial x} (x,t) = \frac{\partial}{\partial x}(ax+bt) \cdot f'(ax+bt)=a f'(ax+bt).$$

Normally, when the context is clear, people usually abuse notation.