Differentiation with different variables

[wilsondd]

Homework Statement

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

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

Homework Equations

FTC will probably be a part of it.

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.

 

[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 $τ$.

 

[wilsondd]

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 $V(t)$.

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

 

[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.

 

[MostlyHarmless]

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.

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

 

[CAF123]

You could perhaps start with the chain rule.

 

[Fredrik]

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?

 

[Ray Vickson]

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

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.

 

[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.