Can you take a derivative with respect to the dependent variable?

[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

 

[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.

 

[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.

 

[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?

Thanks again for your help

 

[dalcde]

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?

Thanks again for your help

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 $\frac{\partial f(x,t)}{\partial t}$).

 

[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

 

[dalcde]

So then does

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

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

 

[micromass]

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

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.

 

[Curious3141]

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.

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:

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

 

[micromass]

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:

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

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

 

[Curious3141]

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

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

 

[micromass]

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

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.

 

[Curious3141]

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.

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.

 

[micromass]

No, sorry, what you wrote is not right.

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.

But seriously, this notation is well-worn and widely used.

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.

 

[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 ) , wherever 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

 

[micromass]

A lot of people hate the Leibiniz notation conventions haha

I sure do

 

[Curious3141]

I sure do

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.

 

[wisvuze]

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.

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

 

[Curious3141]

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

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

 

[micromass]

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

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.

 

[Stalker_VT]

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

But what if there is no equation describing the relation between x and t? As in they are completely independent. Does this not imply that for a differential change in t there would be NO change in x?

i.e.

$\frac{dx}{dt}=0$
Also, micromass and Curious3141, in reference to wheter one can take the derivative of a variable with respect to a function ( i.e. $\frac{dx}{dy}$ ) , if i am understanding you two correctly, Curious3141 is saying the notation is correct, but micromass is saying that it is improper to use that notation because it is senseless?

Thank you

 

[micromass]

Also, micromass and Curious3141, in reference to wheter one can take the derivative of a variable with respect to a function ( i.e. $\frac{dx}{dy}$ ) , if i am understanding you two correctly, Curious3141 is saying the notation is correct, but micromass is saying that it is improper to use that notation because it is senseless?

Let's just say that the notation is correct and widely used, but I hate it.

 

[qbert]

I agree with micromass.

I think the notation is sloppy, which leads to sloppy thinking. it's not much further
along the mathematical evolutionary ladder than the fluctions of Newton.

The good:
Let me give a for instance. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$,
is a given function like $f(x) =\sin(x)$. Then the notation for
$\frac{df}{dx}$ makes some sense
$$\frac{df}{dx}(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}.$$

so if we think of $df \approx \Delta f = f(x+h)-f(x)$ and $dx \approx \Delta x = h$ then
we have something that looks fractional.

The bad:
Then you're invited to write something down without thinking about what you write.
The notation implies $\frac{dx}{df}$ should be the same kind of thing
as $\frac{df}{dx}$. But take a look at
$$\frac{dx}{df}(f) \stackrel{?}{=} \lim_{h \rightarrow 0} \frac{x(f+h) - x(f)}{h}.$$

What does that even mean? if your function happens to be 1-1 at a point
you can give this meaning. This is the meat of the implicit/inverse
function theorems.

But what if f isn't even 1-1?
for my examle, if $f(x) = \sin(x)$ what is
$$\frac{dx}{df}(\pi/3)?$$

 

[Stalker_VT]

for my examle, if $f(x) = \sin(x)$ what is
$$\frac{dx}{df}(\pi/3)?$$

isn't this just 0?

f(x) = $\frac{\pi}{3}$

$\frac{df}{dx}$= 0

$\frac{dx}{df}$= 0

 

[JG89]

^ His point is that if v is a point such that sin(v) = pi/3, then we all know that there is another point, say w such that sin(w) = pi/3. So what is x(pi/3) equal to then? Is it equal to v or is it equal to w? So how can you form the difference quotient [x(pi/3 + h) - x(pi/3)] / h if you cannot assign a unique value to x(pi/3)? You can't.

 

[dalcde]

But what if there is no equation describing the relation between x and t? As in they are completely independent. Does this not imply that for a differential change in t there would be NO change in x?

i.e.

$\frac{dx}{dt}=0$

If they are completely independent, we don't use the total derivative. We simply use the partial derivative (actually the total derivative and the partial derivative are the same when they are independent).

 

[chiro]

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.

I get what you are saying about dx/df in terms of notation, but you can express rates of change in any way you want as long as it makes sense. One instance that this doesn't make sense is when for example a derivative with respect to some variable is 0. The inverse function theorem provides a lot of the clarification for this.

A trivial example is the function y = c. For a two-dimensional cartesian system dy/dx = 0 but dx/dy doesn't make sense for obvious reasons.

In the above case if y = 5x, then there is no problem with saying dx/dy = 1/5 and this is easy to verify on a graph since an inverse function is the reflection of the function against y = x for this particular example and since the relationship is a straight-line, its easy to prove and verify geometrically.