# D and ∂

1. Dec 10, 2013

### Jhenrique

Given f(x(t), y(t)), I know that ∂f/∂x and ∂f/∂y is true (with ∂) because, by definition, use ∂ where f is function of 2 (x, y) or more variable (x, y, z)... ok! But, which theory explain the use of d or ∂ when derive f with respect to parameter t? Is df/dt or ∂f/∂t? And if the function is f(x(t, s), y(t, s)), so, the correct is df/dt or ∂f/∂t? Why? Why? Why?

Thanks!

2. Dec 10, 2013

### Staff: Mentor

Since the first f is a function of two variables, x and y, the derivatives with respect to x and y are partial derivatives: ∂f/∂x and ∂f/∂y.

However, since both x and y are functions of a single variable t, we can also talk about the derivative of f with respect to t: df/dt.

From your earlier threads, the chain rule formula should be familiar to you.
df/dt = ∂f/∂x * dx/dt + ∂f/∂y * dy/dt

Let's call the second function g, to reduce confusion, with g(x(t, s), y(t, s)). Here g is a function of two variables, x and y, but both x and y are functions of two variables, so all derivatives must be partial derivatives.

3. Jan 6, 2014

### Jhenrique

But... if I have $f(\vec{r} (t), t)$, so, f is function of r and t, ie, to derive f with respect to t uses the partial notanion: $\frac{\partial f}{\partial t}$. However, HOWEVER... r is function of t too, so that f is function only of t, ie, the derivative of f with respect to t is $\frac{df}{dt}$, with d. You noticed how the use of d or ∂ is ambiguous!?!?

4. Jan 6, 2014

### D H

Staff Emeritus
There's no ambiguity here. Given your function $f(\vec r(t),t)$, the total derivative of f with respect to time is given by
$$\frac{d f(\vec r(t),t)}{dt} = \frac{\partial f(\vec r(t),t)}{\partial \vec r(t)}\cdot \frac{d\vec r(t)}{dt} + \frac{\partial f(\vec r(t),t)}{\partial t}$$

5. Jan 6, 2014

### PeroK

If you have a function of two variables f(x, y), then for every (fixed) value of y, you have a normal function of x: For each y $f_y(x) = f(x,y)$

This function f_y(x) can be differentiated normally wrt x. So, for each y we have:

$∂f/∂x|_y = df_y/dx$

Or, perhaps clearer:

$\frac{∂f}{∂x} (x, y) = \frac{df_{y}}{dx} (x)$

If both x and y are functions of a third variable t, then f is effectively a function of one variable. In other words, you can plot a normal 2-D graph of f(t) and differentiate it normally.

And, if you have f(x(t), t), then you cannot have a partial derivative wrt t, because you cannot fix x and then differentiate wrt t. As t varies, both variables vary.

6. Jan 6, 2014

### D H

Staff Emeritus
Nonsense. The partial of f with respect to t ignores that the other argument x is also a function of time.

7. Jan 6, 2014

### Jhenrique

it's confuse...

8. Jan 6, 2014

### PeroK

Yes, of course, you can go through the formal process of partial differentiation, but it's not really the point. Once you know f is a function of t as a single variable, it's the regular derivative you're after.

Partial derivatives really apply when f is a function of two or more free variables.

Last edited: Jan 6, 2014
9. Jan 6, 2014

### pwsnafu

Here is how I would approach the problem of $f(r(t),t)$. First we have $f$ with respect to two variables, which I'll call x and y. These are dummy variables. We have $f(x,y)$.

However, x depends on another variable t. Hence $x = r(t)$.
Similarly, y depends on t also. But the trick is that this is actually the identity function $y = Id(t)$. Some students just like to write this as $y = t$.

Now the chain rules says
$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$.

So firstly what is $\frac{dx}{dt}$? Well $x = r(t)$ so $\frac{dx}{dt} = \frac{dr}{dt}$.
Now the other one. What is $\frac{dy}{dt}$? Ah, but you see the derivative of the identity function is 1. So we get $\frac{dy}{dt} = 1$.

We now write the answer out exactly:
$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}$.
Oh but wait! $x = r(t)$ and $y = t$ remember. We should correct that:
$\frac{df}{dt} = \frac{\partial f}{\partial r}\frac{dr}{dt} + \frac{\partial f}{\partial t}$.

10. Jan 6, 2014

### PeroK

... just to add a note to what pwsnafu has said. We can define:

$g:\mathbb{R} → \mathbb{R} \ s.t. \ g(t) = f(x(t), y(t))$

And then, strictly speaking, it's g that we diffentiate wrt t using the partials of f wrt x and y, the chain rule and the normal derivatives of x and y wrt t.

And that sorts things out in terms of always differentiating a well-defined function wrt the "correct" variables.

11. Jan 6, 2014

### Office_Shredder

Staff Emeritus
PeroK, that is formally a good way of defining things but in practice everyone will simplify the notation and write $\frac{\partial f}{\partial t}$ when talking about f(x(t),t).

12. Jan 6, 2014

### PeroK

Yes, I agree, but hopefully it helps the OP see where to use ∂ and d and understand the difference.

13. Jan 7, 2014

### arildno

If you want to be REALLY careful when you meet an uglyargument function, say f(x(t,s),t), you should introduce auxiliary functions with their own, distinct names. In that way, you won't get confused when differentiating.

In the above, the fundamental function is f(x,t). Also define the function X(t,s).
Then, we introduce a THIRD function, F(t,s), defined by the identity F(t,s)=f(X(t,s),t)

Here, we have:
$$\frac{\partial{F}}{\partial{t}}=\frac{\partial{f}} {\partial{x}}\frac{\partial{X}}{\partial{t}}+ \frac{\partial{f}}{\partial {t}}$$

Note that the last addend here is perfectly well defined, because we have made the explicit definition that f has two INDEPENDENT variables, "x" and "t". The definition of F as an explicitly different function, shows why its partial derivative with respect to "t" is totally different from f's partial derivative with respect to "t".
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As you can see, I have a strong sympathy for Perok's formal approach; for truly uglynastinesses, it is the simplest approach, although it is overly tedious for the simplest examples.

Last edited: Jan 7, 2014
14. Jan 7, 2014

### vanhees71

In such cases it's sometimes better to put the arguments explicitly. So here, I'd write
$$\partial_t f[X(t,s),t]=\left [\partial_x f(x,t) \right ]_{x=X(t,s)} \partial_t X(t,s) + \left [\partial_t f(x,t) \right]_{x=X(t,s)}.$$

15. Jan 7, 2014

### arildno

Absolutely. That's the next level (or alternate manner) of elucidating specification.
But, writing first down the table of the auxiliary functions you are using, and how they are related remains an alternative.