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If I have the function $ h \equiv h(f(x,t),g(x,t))$, where $x$ and $t$ are independent variables and $f$ and $g$ are functions of $x$ and $t$.

Then is,

$$\frac{\mathrm{d}h }{\mathrm{d} t} = h\left(\frac{\mathrm{d}f }{\mathrm{d} t},g \right) + h\left(f , \frac{\mathrm{d}g }{\mathrm{d} t}\right)$$

or

$$\frac{\mathrm{d}h }{\mathrm{d} t} = h\left(\frac{\partial{d}f }{\partial{d} t},g \right) + h\left(f , \frac{\partial{d}g }{\partial{d} t}\right)\ ?$$

And can you tell me why? I think the first one might be correct.

Matrix23
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2 Answers2

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This is just chain rule:

$$\frac{\partial h}{\partial t}(f(x,t),g(x,t)) = \frac{\partial h}{\partial f}(f(x,t),g(x,t))\frac{\partial f}{\partial t}(x,t) + \frac{\partial h}{\partial g}(f(x,t),g(x,t))\frac{\partial g}{\partial t}(x,t).$$

DMcMor
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    Please keep in mind I'm asking for $\frac{\mathrm{d}h }{\mathrm{d} t}$ – Matrix23 Jul 17 '17 at 21:17
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    What do you mean by $\frac{\mathrm{d}h }{\mathrm{d} t}$? Since $h$ is a function of both $x$ and $t$ you need to work with partial derivatives. – DMcMor Jul 17 '17 at 21:19
  • From a physical perspective, say a particle moves in the $x-y$ plane. There is a field $f \equiv f(x,y)$ defined on the plane. Say, my particle moves along a straight line trajectory and arrives at a point in the $x-y$ plane. Consider two cases, First case, particle is moving parallel to the $x$-axis. Second case, particle is moving at a $45^{o}$ angle to the x-axis. I can ask the question: for a unit change in the value of $x$-coordinate of the particle, what is the corresponding change in $f$. I believe the answer is potentially different in the two cases. – Matrix23 Jul 17 '17 at 22:14
  • ${\frac{\partial{f} }{\partial{x} }}$ gives only one answer at a given point on the plane. However,$\frac{\mathrm{d}f }{\mathrm{d} x} = \frac{\partial{f} }{\partial{x} } + \frac{\partial{f} }{\partial{y} } \frac{\mathrm{d}y }{\mathrm{d} x}$. It is $\frac{\mathrm{d}y }{\mathrm{d} x}$ that takes into account the instantaneous direction of trajectory at that point on the plane. – Matrix23 Jul 17 '17 at 22:19
  • Thinking about this a bit more.. In the case of a particle trajectory, $x$ and $y$ are not independent.... So,$ \frac{\mathrm{d}f }{\mathrm{d} x}$ might be valid. – Matrix23 Jul 18 '17 at 00:07
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Since I hate this abuse of notation, I'll introduce new variables

  • $u = f(x,t)$
  • $v = g(x,t)$
  • $y = h(u,v)$

Now, back to the question.

Except in degenerate cases, $\frac{\mathrm{d}y}{\mathrm{d}t}$ doesn't exist at all!

The differential of $y$ is given by

$$\begin{align} \mathrm{d}y &= h_1(u,v) \mathrm{d}u + h_2(u,v) \mathrm{d}v \\ &= h_1(u,v) \left( f_1(x,t) \mathrm{d}x + f_2(x,t) \mathrm{d}t \right) + h_2(u,v) \left( g_1(x,t) \mathrm{d}x + g_2(x,t) \mathrm{d}t \right) \\&= \left( h_1(u, v) f_1(x,t) + h_2(u,v) g_1(x,t) \right) \mathrm{d}x + \left( h_1(u, v) f_2(x,t) + h_2(u,v) g_2(x,t) \right) \mathrm{d}t \end{align}$$

where I've used subscript notation on functions to refer to which place to take the derivative of a function.

The premise that $x$ and $t$ are independent implies that $\mathrm{d}x$ and $\mathrm{d}t$ are linearly independent. Thus, $\mathrm{d}y$ can only be a multiple of $\mathrm{d}t$ at those places where

$$ h_1(u, v) f_1(x,t) + h_2(u,v) g_1(x,t) = 0 $$

On any domain where this doesn't hold, the notion of $\frac{\mathrm{d}y}{\mathrm{d}t}$ is simply nonsensical.

For those who prefer the following notation style, the above formula would usually be written as

$$ \mathrm{d}h = \left( \frac{\partial h}{\partial f} \frac{\partial f}{\partial x} + \frac{\partial h}{\partial g} \frac{\partial g}{\partial x} \right) \mathrm{d}x + \left( \frac{\partial h}{\partial f} \frac{\partial f}{\partial t} + \frac{\partial h}{\partial g} \frac{\partial g}{\partial t} \right) \mathrm{d}t $$

and the condition needed for $\frac{\mathrm{d}h}{\mathrm{d}t}$ to exist is

$$ \frac{\partial h}{\partial f} \frac{\partial f}{\partial x} + \frac{\partial h}{\partial g} \frac{\partial g}{\partial x} = 0 $$

  • For a simple case where a particle moves in the $x$-$y$ plane and there is a field $f \equiv f(x,y)$ defined on the plane, we can talk about the trajectory of a particle in the $x$-$y$ plane. In this case,$y$ is function of $x$. So, can we always talk about $\frac{\mathrm{d}f }{\mathrm{d} x}$ here? Alternatively, in the case that I had provided in the original question where $x$ and $t$ are independent of each other, can we talk about $\frac{\mathrm{d} x}{\mathrm{d} t}$? – Matrix23 Jul 17 '17 at 23:53
  • @IanDsouza: If $\mathrm{d}y$ is a multiple of $\mathrm{d}x$ and $\mathrm{d}f$ is a linear combination of $\mathrm{d}x$ and $\mathrm{d}y$, then by substitution $\mathrm{d}f$ is a multiple of $\mathrm{d}x$, so in that case $\frac{\mathrm{d}f}{\mathrm{d}x}$ would make sense. If $x$ and $t$ are independent, $\mathrm{d}x$ is nowhere a multiple of $\mathrm{d}t$, so $\frac{\mathrm{d}x}{\mathrm{d}t}$ wouldn't make sense. –  Jul 18 '17 at 03:49