# Multivariable analog to the total derivative?

1. Jun 17, 2010

### pellman

For a single variable we have

$$\int_{x_1}^{x_2} f(x) dx = F(x_2)-F(x_1)$$

if f(x) = dF/dx. f(x) is then a total derivative. What is the analog in 3D so that

$$\int_V f(\vec{x}) d^3x$$

does not depend on the values of f in the interior of V?

In case there is not a single answer, let me give the context. In the calculus of variations two Lagrangians are equivalent if

$$L_2(q(t),\dot{q}(t),t)=\lambda L_1(q(t),\dot{q}(t),t) + \frac{d}{dt}F(q(t),\dot{q}(t),t)$$

where lambda is a constant and F is any function. (That is, their actions are extremized for the same function q(t).) What replaces dF/dt in this equivalency if we have a multi-parameter action

$$S=\int L(q(\vec{x}),\partial q(\vec{x}),\vec{x}) d^3x$$

(where $$\partial q$$ stands for the various partial derivatives of q)?

Is it $$\nabla \cdot \vec{F}$$ for some vector function F? Or is there more to it than that?

Last edited: Jun 17, 2010
2. Jun 17, 2010

### Staff: Mentor

I think you have your terminology wrong. In your first example F is an antiderivative of f and f is the derivative of F.

The total derivative refers to a function of two or more variables, for example f(x, y). The total differential of f in this case is
$$df = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy$$

If it turns out that x and y are differentiable functions of t, then the total derivative of f looks like this:
$$\frac{df}{dt} = \frac{\partial f}{\partial x}~\frac{dx}{dt} + \frac{\partial f}{\partial y}~\frac{dy}{dt}$$

3. Jun 17, 2010

### l'Hôpital

4. Jun 18, 2010

### pellman

Thanks to both .

5. Jun 18, 2010

### Studiot

6. Jun 19, 2010

### pellman

Thanks, Studiot