# How to differentiate this potential energy function?

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1. Aug 15, 2015

### diredragon

1. The problem statement, all variables and given/known data
Given the potential energy function V(x,y)=V(ax-by) where a,b is an arbitrary constants differentiate with respect to x and y.

2. Relevant equations
Multivariavle differentiation

3. The attempt at a solution
(d/dt)p2=+bV'(ax-by). The right side is confusing. How did it get to that? Im not in my MultiCal class yet as im just in the middle of SingCal so i dont get it. Can you guide through the process?

2. Aug 15, 2015

### ShayanJ

Assuming $z=ax-by$ and using chain rule, we have:
$\frac{dV}{dx}=\frac{dV}{dz}\frac{dz}{dx}=a\frac{dV}{dz}$. Now we change the name of $\frac{dV}{dz}$ to $V'$.

3. Aug 15, 2015

### diredragon

Ok, but why is it -aV'?

4. Aug 15, 2015

### ShayanJ

At first I didn't know what are p1 and p2 and what are those equations, but I thought you yourself know! Anyway, I think those are the x and y components of Newton's 2nd law and so they are components of $\frac{d\vec p}{dt}=-[\frac{dV}{dx}\hat x+\frac{dV}{dy}\hat y]$. The extra minus sign is because of the minus sign in the formula itself.

5. Aug 15, 2015

### diredragon

Oh yes, i know what the equations are but have totally forgotten about the minus in front of the right side of lagrangian. Thanks