How to differentiate this potential energy function?

[diredragon]

Homework Statement

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.

Homework Equations

Multivariavle differentiation

The Attempt at a Solution

The answer yields (d/dt)p1=-aV'(ax-by)
(d/dt)p2=+bV'(ax-by). The right side is confusing. How did it get to that? I am not in my MultiCal class yet as I am just in the middle of SingCal so i don't get it. Can you guide through the process?

 

[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 $V'$.

 

[diredragon]

Ok, but why is it -aV'?

 

[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.

 

[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