# Differentiation with different variables

1. Mar 4, 2013

### wilsondd

1. The problem statement, all variables and given/known data

I'm trying to take the derivative of the following integral

$\frac{d}{d V} \int_0^t{V(\tau)}d\tau$

2. Relevant equations

FTC will probably be a part of it.

3. The attempt at a solution

I always get confused when I'm taking the derivative of an integral. I know the answer is not simply $V(t)$ though since the integrating variable is not the same as the variable I am trying to take the derivative with respect to.

2. Mar 4, 2013

### hapefish

I found the expression $\frac{d}{d V} \int_0^t{V(\tau)}d\tau$ a little surprising. I'm not sure how to interpret the $\frac{d}{d V}$. I would have expected to see $\frac{d}{d t} \int_0^t{V(\tau)}d\tau$ instead. In that case it is exactly the fundamental theorem of calculus and the solution is $V(t)$.

Note that I would expect the solution to be a function of $t$ because the expression itself is a function of $t$. $τ$ is a dummy variable used to compute the definite integral - the expression itself is not a function of $τ$.

3. Mar 4, 2013

### wilsondd

Yes, I agree it is weird, but it is part of a calculus of variations problem, and is not a typo or anything.

4. Mar 4, 2013

### wilsondd

Thinking more about the problem, if it is legal to bring the derivative inside the integral, then the answer would just be $t$. This seems odd in the context of the larger problem, though.

Can anybody comment on if this is correct or not. Thanks.

5. Mar 4, 2013

### MostlyHarmless

That is not legal. The only thing you can move in and out of integrals is contants.

6. Mar 4, 2013

7. Mar 4, 2013

### Fredrik

Staff Emeritus
You may need to provide more information. What does d/dV even mean here? It can't be a derivative, since the thing to the right of it isn't the value at V of some differentiable function. Is it a functional derivative? Does the course you're taking cover the definition of "functional derivative"? Does the book prove a theorem for functional derivatives that is similar to the chain rule?

8. Mar 4, 2013

### Ray Vickson

In Calculus of variations and Optimal Control we talk about "functional derivatives", which are like directional derivatives in function space. So, given a functional
$$F(f) = \int_a^b L(f(t)) \, dt,$$
the directional derivative of F(f) in the direction h (also called the Gateaux derivative) is
$$DF(f;h) \equiv \frac{d}{d\epsilon} \int_a^b F(f(t) + \epsilon h(t)) \, dt\\ = \int_a^b F^{\prime}(f(t))\cdot h(t) \, dt.$$
In this notation, your functional $F(V) = \int_0^t V(\tau) \, d\tau$ would have directional derivative
$$DF(V;h) = \int_0^t h(\tau) \, d\tau.$$

I don't know if that is what you want, but that is what is done in all treatments of Calculus of Variations that I have seen.

9. Mar 4, 2013

### wilsondd

$\frac{d}{dV}$ is the derivative with respect to the function $V(t)$.

This is a calculus of variations problem, and the functional is

$\int_0^{t_1}{\mathcal{M}}dt$,

$\mathcal{M} = \int_0^t{V(\tau)}d\tau + \text{other stuff...}$

What I'm trying to do, is find the equation for

$\frac{d\mathcal{M}}{dV} = \frac{d}{dt} \frac{d\mathcal{M}}{d\dot{V}}$

which is where the derivative comes from in the first place.

Hope this helps.

Thanks.