# Differentiation Quesion

1. Jan 10, 2012

### Stalker_VT

If you are given an equation y(x), where y is the dependent variable, and x is the independent variable, can you take a derivative with respect to the dependent variable?

ex.

y = 5x

dy/dx = 5

dx/dy = 1/5

Thank you for any help

2. Jan 10, 2012

### zhermes

Absolutely, that definitely works.
From a philosophical standpoint, the mathematics doesn't know which is the dependent and which is the independent variable---anything you can do to one, you can do to the other.

3. Jan 10, 2012

### dalcde

Just be careful to make sure that it makes sense:
You can take the derivative of position with respect to time and get the velocity, which is how fast your position changes per unit time, but the derivative of time with respect to position makes less sense.

4. Jan 10, 2012

### Stalker_VT

That's what i thought. Thanks!

Now what if you had an equation of multiple independent variables such as f(x,t) and you take the Total Derivative of f(x,t) with respect to t...using the chain rule

$\frac{D f(x,t)}{D t}$ = $\frac{\partial f(x,t)}{\partial x }$$\frac{dx}{dt}$ + $\frac{\partial f(x,t)}{\partial t}$$\frac{dt}{dt}$

my quesions are

1) $\frac{dt}{dt}$ = 1 ? (i think this is obvious)

2) $\frac{dx}{dt}$ = 0 ? because both x and y are independent of one another?

5. Jan 10, 2012

### dalcde

The total derivative is used when the other variables also depend on t. So $\frac{dx}{dt}\not=0$. Or else we can just use the partial derivative. It's true that $\frac{dt}{dt}$ = 1. If you check the Wikipedia page, it doesn't even include that (we only have $\frac{\partial f(x,t)}{\partial t}$).

6. Jan 10, 2012

### Stalker_VT

So then does

$\frac{dx}{dt}$ = 1?

meaning that $\frac{D f(x,t)}{D t}$ = $\frac{\partial f(x,t)}{\partial x }$ + $\frac{\partial f(x,t)}{\partial t}$

Thank you

7. Jan 11, 2012

### dalcde

No. You treat it as a normal derivative (so it can be anything). For example, if $x=t^2+6$, then $\frac{dx}{dt}=2t$.

8. Jan 11, 2012

### micromass

AAAAAAAAAAAAAAAAAAAAAAAAARRRRRRRRRRGGGGGGGGGGHHHHHHHHHH. *faints*

No, you can't do anything like that. y is a function, that is: y is defined as $y:\mathbb{R}\rightarrow \mathbb{R}$ and x is defined as an element of $\mathbb{R}$.

So dy/dx is just a notation for the derivative of the function y with respect to x.
Writing dx/dy has no sense what-so-ever. That is supposed to mean the derivative of a variable with respect to a function. This is ill-defined. Do not write anything like that.

9. Jan 11, 2012

### Curious3141

Actually, I thought it was perfectly sound notation to write $\frac{dx}{dy}$. The proof of the "reciprocal" thing is a simple consequence of the Chain Rule:

$\frac{dy}{dx} = \frac{\frac{dy}{dy}}{\frac{dx}{dy}} = \frac{1}{\frac{dx}{dy}}$.

10. Jan 11, 2012

### micromass

But x isn't even a function!! How can we find the derivative of something that isn't even a function??

11. Jan 11, 2012

### Curious3141

$x = f^{-1}(y)$, provided the inverse function is well-defined and differentiable.

12. Jan 11, 2012

### micromass

That doesn't mean that x is a function. That only means that x is a real number.

That said, it IS true that

$$\frac{df}{dx}=\frac{1}{\frac{df^{-1}}{dx}}$$

This is probably what you mean.

13. Jan 11, 2012

### Curious3141

No, sorry, what you wrote is not right.

Let's say $y = f(x)$. Then $x = f^{-1}(y) = g(y)$ (in other words, we rename the inverse function of f as g for clarity).

What you wrote is: $f'(x) = \frac{1}{g'(x)}$ which is NOT correct.

The correct form is: $\frac{dy}{dx} = f'(x) = \frac{1}{\frac{dx}{dy}} = \frac{1}{g'(y)} = \frac{1}{g'(f(x))}$.

A simple illustration. Let $y = f(x) = x^2$, where $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$. Then $g(x) = f^{-1}(x) = \sqrt{x}$, $g:\mathbb{R}^+ \rightarrow \mathbb{R}^+$.

$f'(x) = 2x$, $g'(x) = \frac{1}{2}x^{-\frac{1}{2}}$

What you're asserting is:

$2x = \frac{1}{\frac{1}{2}x^{-\frac{1}{2}}} = 2\sqrt{x}$, which is obviously false.

What I'm asserting is:

$2x = \frac{1}{\frac{1}{2}y^{-\frac{1}{2}}} = 2\sqrt{y} = 2x$, which is obviously true.

Not a proof, just an illustration. But seriously, this notation is well-worn and widely used.

14. Jan 11, 2012

### micromass

Indeed not; it was nonsense. What I should have written was

$$\frac{df^{-1}}{dx}(f(a))=\frac{1}{\frac{df}{dx}(a)}$$

Or in the multivariable version:

$$D f^{-1}(f(a))=(Df(a))^{-1}$$

This is exactly the so-called inverse function theorem.

I don't really care. The notation is wrong. It doesn't make any sense.

A notation $\frac{df}{dx}$ means that f is a function and that x is a dummy variable. Writing it as $\frac{dx}{df}$ makes no sense.

15. Jan 11, 2012

### wisvuze

Yeah, the dy/dx notation ( or dt/dy ) comes up in differential equations quite a bit ( as it would be a useful computational piece of notation I guess ) , whereever you can actually express x as a function of y ( i.e. inverse function is well-defined ).

A lot of people hate the Leibiniz notation conventions haha

16. Jan 11, 2012

### micromass

I sure do

17. Jan 11, 2012

### Curious3141

Doesn't mean it's wrong! :tongue:

I was about to quote the wiki on the Leibniz notation (oh heck, here it is: http://en.wikipedia.org/wiki/Differentiation_rules#The_inverse_function_rule)), but I guess you already know it and deplore it in advance.:rofl:

What we really need is a professional mathematician to chip in here (unless you're one) and resolve this.

18. Jan 11, 2012

### wisvuze

I think it's been settled that the notation is in well-use. My professor last semester used it all the time. Anyway, notation is just notation, whether or not that specific piece of leibiniz notation is all a matter of naming things in a way you want everything to make sense

19. Jan 11, 2012

### Curious3141

OK, thanks. I've seen this used countless times, in reputable advanced reference texts on Analysis, too. I never had an issue with it.

20. Jan 11, 2012

### micromass

The notation is certainly used a lot. That doesn't mean that I don't consider it to be wrong.

I try to avoid Analysis books that use Leibniz notation because I find them not rigorous enough.