# Homework Help: Understanding derivation of -du/dx = F

1. Oct 2, 2016

### hsbhsb

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

Where U is potential energy, show that F = -du/dx

2. Relevant equations

ΔU = U(x) - U(x_0)

$$ΔU = U(x) - U(x_0) = \int_{x_0}^{x}Fdx \qquad (1)$$

3. The attempt at a solution

I am confused why
$$\frac{d}{dx}ΔU = \frac{du}{dx} \qquad (2)$$

Doesn't
$$ΔU = du \qquad (3)$$

I recognize that (3) is simple misunderstanding of calculus but for some reason I can't make sense of why it isn't true in this case.

2. Oct 2, 2016

### TSny

I'm a little confused with the notation. Is $u$ the same as $U$?

Note that $\frac{d}{dx}ΔU = \frac{d}{dx} \left[U(x) - U(x_0) \right]$.

Simplify this.
Hint: What is $\frac{d}{dx} U(x_0)$?

Why do you think it is true in this case?

3. Oct 3, 2016

### hsbhsb

Yes it is meant to, my mistake for lack of clarity. From now on I will use $dU$ instead of $du$ to represent the derivative of $U$
Does it $=F+dU/dx$?

Ok, I can see why $ΔU \neq dU$ in this case. After all, $ΔU$ represents a real change in potential, not an infinitesimally small one. But I am still having trouble understanding this algebraically and intuitively. How can you take the derivative of $ΔU$? In a situation where all potential energy is provided by gravity, $ΔU$ does not represent a single point where gravity has a single instantaneous force, like $U(x)$ and $U(x_0)$ do. It is rather, a range. Does taking the derivative of $ΔU$ then yield an average force over that range?

I think my grasp of Leibniz notation is poor.

4. Oct 3, 2016

### TSny

You have the relation $$ΔU = U(x) - U(x_0) = \int_{x_0}^{x}Fdx$$ I believe there should be a negative sign in this relation, so that it should read $$ΔU = U(x) - U(x_0) = - \int_{x_0}^{x}Fdx$$ Think of $x_0$ as having a fixed value; but consider $x$ to be a variable. So, $U(x_0)$ is just some number (i.e., a constant), but $U(x)$ is a function of the variable $x$.

Use rules of calculus to simplify $\frac{d}{dx}ΔU = \frac{d}{dx}\left[ U(x) - U(x_0) \right] = -\frac{d}{dx}\int_{x_0}^{x}Fdx$

5. Oct 3, 2016

### hsbhsb

I get it! Thank you :)