# Simple question regarding Product Rule

#### bitrex

1. The problem statement, all variables and given/known data
As part of a first order differential equation I need to find $$\frac{d}{dt}(mv)$$
where $$v=\frac{dy}{dt}$$

2. Relevant equations

Product Rule.

3. The attempt at a solution

$$\frac{d}{dt}(m\frac{dy}{dt}) = m*\frac{d}{dt}\frac{dy}{dt} + \frac{dm}{dt}\frac{dy}{dt} = ??$$

I know I shouldn't have to deal with a second derivative, at least in this equation. Is there a way to simplify the equation first that I'm not seeing?

Related Calculus and Beyond Homework News on Phys.org

#### Dick

Homework Helper
Why do you think there's no second derivative? There is and you did it correctly. The second derivative would be the acceleration. If m*v represents a momentum you can often take the mass m to be constant. That would simplify it.

#### bitrex

Well, the equation I'm dealing with in full is $$\frac{d}{dt}(mv) = {mg }$$, where m is a function $$4/3{\pi}kt^3$$. Ahh, I see how to set the equation up. After I get $$m\frac{d^2y}{d^2t} + \frac{dm}{dt}\frac{dy}{dt}$$ I substitute $$\frac{dv}{dt}$$ for $$\frac{d^2y}{d^2t}$$ and v for $$\frac{dy}{dt}$$. I then solve the differential equation for v(t) with initial condition v(0) = 0. I have to prove that in this case $$\frac{d^2y}{d^2t}$$ is proportional to $$\frac{g}{4}$$, so once I know velocity I can differentiate the function to get acceleration and see if that works out. I think I can do this now - I just needed my brain jogged a bit. Thanks!

"Simple question regarding Product Rule"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving