# In equation (3) from lecture 7 in Leonard Susskind’s ‘Classical Mechanics’, should the derivatives be partial?

Here are the equations. ($$V$$ represents a potential function and $$p$$ represents momentum.) $$V(q_1,q_2) = V(aq_1 - bq_2)$$ $$\dot{p}_1 = -aV'(aq_1 - bq_2)$$ $$\dot{p}_2 = +bV'(aq_1 - bq_2)$$

Should the $$V’$$ terms be partial derivatives (with respect to $$q_1$$ and $$q_2$$ for $$\dot{p}_{1}$$ and $$\dot{p}_{2}$$, respectively?).

To help clarify, it might help to use different letters for the potential in this equation: $$V(q_1, q_2) = f(a q_1 - b q_2)$$ In other words, on the left hand side, $$V$$ is a function of two variables, while on the right hand side, $$f$$ is a function of one variable. Susskind uses the same letter for both functions (which is a common step in physics), but note that they are strictly different mathematical objects.
The other thing to remember is what the notation $$f'(a q_1 - b q_2)$$ means. It means to first evaluate the derivative of $$f$$ with respect to its argument, then plug in the value $$a q_1 - b q_2$$. In other words, $$f'(a q_1 - b q_2) = \frac{df}{dx}\Big|_{x=a q_1 - b q_2}$$
Given these notational clarifications, I think it should hopefully be easy to see that, for example, $$p_1 = - \frac{\partial V}{\partial q_1} = - \frac{d f}{dx} \frac{\partial x}{\partial q_1} = - a f'(a q_1 - b q_2)$$ where $$x=a q_1 - b q_2$$.
They are partial derivatives. From the chain rule, we have $$\frac{\partial V(aq_1-bq_2)}{\partial q_1}= a V'(aq_1-bq_2),\\ \frac{\partial V(aq_1-bq_2)}{\partial q_2}= -b V'(aq_1-bq_2).$$ For example $$\frac{\partial \sin(aq_1-bq_2)}{\partial q_1}= a \cos(aq_1-bq_2),\\ \frac{\partial \sin (aq_1-bq_2)}{\partial q_2}= -b \cos(aq_1-bq_2).$$ Here, $$V(x)= \sin(x)$$ and $$V'(x)=\cos(x)$$.