# Find the derivative of a function

1. Jun 23, 2014

### mohams

1. The problem statement, all variables and given/known data
If V=exp [ $\int$$^{T}_{0}$s(t)dt ]

2. Relevant equations
What is dV/ds(k), where 0<k<T
What does this derivative even mean??

3. The attempt at a solution
write
V=exp(Y)
dV/ds(k) = dV/dY . dY/ds(k)
=V.$\int$$^{T}_{0}$ds(t)/ds(k)dt
=V because ds(t)/ds(k) = 0 for all t except t=k where it is 1.

I just want to check that this derivative is correct; and more importantly what it means practically. Thank you very much for the help!

Regards,

Last edited: Jun 23, 2014
2. Jun 24, 2014

### mohams

Hello all,
I've been thinking about the concept of the functional derivative all night/day but I still can not see the practical meaning of such a measure. Scalar derivative - no problem. It is simply the rate of change wrt the scalar variable. But when you have a function argument, and you take the rate of change with regards to just a point on that function argument, then what are you measuring exactly? Any help much appreciated! Thanks!

3. Jun 25, 2014

### bloby

A way to look at it:
s, f, g $\in$ M the function space; $V:M\to \mathbb R$ a functional.

- $< f, g> = \int f(x)g(x)\,dx$ is a bilinear, symetric and "almost" non-degenerate form. It associate a fuction with a form: $$f\in M \mapsto <f,\cdot> \,or\,<\cdot,f>\in M^*$$

- The Dirac delta function $\delta (x) = \delta_0$ or $\delta(x-y) = \delta_y$ $\in$ M, y fixed, is the function that correspond to the form $$<\delta_y,\cdot>\in M^*\quad s.t. \quad <\delta_y,f> = f(y)\in \mathbb R$$

- The functional derivative of a functional at the point(=function) $g\in M$ is the function $\frac{\delta V}{\delta g}$ or $\frac{\delta V}{\delta g(x)}$ (as the notations f or f(x)) such that: $$<\frac{\delta V}{\delta g},\cdot>\in M^* = \mathbf D_g V(\cdot)$$ $D_g V(\cdot)$ the derivative of V at g. It's a linear map sending a vector h at g (h=function again) to the directional derivative of V at g in the direction h:$$D_g V(\cdot): h\mapsto D_g V(h) = \lim_{\epsilon \rightarrow 0}{\frac{V(g+\epsilon h)-V(g)}{\epsilon}}\in \mathbb R$$

Then $<\delta_y,\frac{\delta V}{\delta g}> = \frac{\delta V}{\delta g(y)}$ the value of the function $\frac{\delta V}{\delta g(x)}$ at x=y.