rakuenso Posted May 14, 2009 Share Posted May 14, 2009 Hi I have a function f: [math]f(x(t))=\frac{d(x(t))}{dt}+x(t)[/math] Now how would I differentiate f with respect to x Link to comment Share on other sites More sharing options...

Royston Posted May 14, 2009 Share Posted May 14, 2009 My initial idea was use the chain rule, have you tried solving it with a matrix ? Try separating it with a matrix, you should end up with two differential equations. Link to comment Share on other sites More sharing options...

Bignose Posted May 14, 2009 Share Posted May 14, 2009 Assuming all the function are well-behaved (i.e. all the derivatives exist) you just take the derivative with respect to x straight forwardly. I'm going to drop the (t) notation -- it is still in there, x is still a function of t, I'm just not going to write it all out. [math]\frac{d}{dx} f = \frac{d}{dx} \left(\frac{dx}{dt} +x\right)[/math] [math]\frac{d}{dx} f = \frac{d}{dx}\frac{dx}{dt}+\frac{d}{dx}x[/math] now, this is where the "well-behaved" comes in, the order of the differentiation with respect to x & t can be reversed: [math]\frac{d}{dx} f = \frac{d}{dt}\frac{dx}{dx}+\frac{d}{dx}x[/math] [math]\frac{d}{dx} f = \frac{d}{dt}1+1[/math] [math]\frac{d}{dx} f = 1[/math] I have no idea what Snail is talking about with matrices and the like. What matrix? where? Furthermore, what is there to "solve"? I really don't get any of Snail's reply. Link to comment Share on other sites More sharing options...

Royston Posted May 14, 2009 Share Posted May 14, 2009 I was replying to a completely different forum and question, not sure how that happened....sorry about that. Link to comment Share on other sites More sharing options...

Bignose Posted May 14, 2009 Share Posted May 14, 2009 I was replying to a completely different forum and question, not sure how that happened....sorry about that. That makes a lot more sense -- I can move posts around the math section here, but I don't think they've given me enough power to move posts across forums (...yet). Link to comment Share on other sites More sharing options...

rakuenso Posted May 15, 2009 Author Share Posted May 15, 2009 Bignose, under what conditions is [math]\frac{d}{dt}\frac{dx}{dx}=\frac{d}{dx}\frac{dx}{dt}[/math]? I tried to do some examples, say: [math] x = t^2 [/math] [math] \frac{dx}{dt} = 2t [/math] [math]\Rightarrow[/math] [math] \frac{d}{dx}\frac{dx}{dt} = \frac{1}{\sqrt{x}}[/math] since [math] \sqrt{x}=t [/math] However, [math] \frac{d}{dt}\frac{dx}{dx} = 0 [/math] So [math] \frac{d}{dt}\frac{dx}{dx}\neq \frac{d}{dx}\frac{dx}{dt} [/math] Link to comment Share on other sites More sharing options...

Bignose Posted May 15, 2009 Share Posted May 15, 2009 rak, I'm not ignoring your question, I just haven't had a lot of time to get back to it. My first instinct is that because your f(x) actually isn't a function (which maps a single number to a single number) but is actually a functional (which maps a function to a function), that the definition of differentiation of the function isn't the quite right. see http://mathworld.wolfram.com/FunctionalDerivative.html I need some more time to look at this, so I'll get back when I've thought about it more... Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now