# Simple question regarding Product Rule

• bitrex
In summary, the conversation is discussing finding the derivative of momentum (mv) in a first order differential equation where v is equal to the derivative of y with respect to t. The product rule is used and the conversation also touches on dealing with a second derivative and simplifying the equation by assuming the mass is constant. The final equation being worked on is \frac{d}{dt}(mv) = {mg }, and the conversation ends with a plan to solve for v(t) and prove the relationship between acceleration and gravity.
bitrex

## Homework Statement

As part of a first order differential equation I need to find $$\frac{d}{dt}(mv)$$
where $$v=\frac{dy}{dt}$$

Product Rule.

## 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?

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.

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!

## 1. What is the product rule in calculus?

The product rule is a formula used in calculus to find the derivative of a product of two functions. It states that the derivative of the product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function.

## 2. How do I use the product rule to find the derivative of a function?

To use the product rule, you first identify the two functions being multiplied together. Then, you take the derivative of each function separately. Finally, you plug in these values into the product rule formula to find the derivative of the product of the two functions.

## 3. Why is the product rule important in calculus?

The product rule is important because it allows us to find the derivative of a product of two functions, which is a common occurrence in many mathematical and scientific applications. It also helps us to simplify more complex functions and solve problems involving rates of change.

## 4. Can the product rule be applied to more than two functions?

Yes, the product rule can be extended to find the derivative of a product of any number of functions. The general formula for the product rule is f'(x)g(x) + f(x)g'(x) + h(x)i(x)j(x) + h'(x)i(x)j(x) + h(x)i'(x)j(x) + h(x)i(x)j'(x) + ... and so on.

## 5. Are there any special cases or exceptions to the product rule?

Yes, there are some special cases where the product rule may not apply. For example, if one of the functions is a constant, then the derivative of that function will be zero and the product rule will not be necessary. Also, if the two functions are the same, then the product rule simplifies to the power rule.

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