Get Expert Tips for Proving the Product Rule Using Gradient Method

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In summary, the problem is verifying the product rule using the gradient of two functions. The rest of the problem is on the backside file, but help getting started on the first part would be enough. Any help would be greatly appreciated.
  • #1
Fjolvar
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I've been up for hours trying to figure out the proper notation to prove problem #4 (1.6.5) on the attached files. I'm not sure how to set this up. It is basically proving the product rule using the gradient of two functions. The rest of the problem is on the backside file, but help getting started on the first part would be enough. Any help would be greatly appreciated. Thanks!
 

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  • #2
Any ideas?
 
  • #3
It looks to me like a pretty standard "just plug it in and calculate"!

You are asked to show that [itex]\nabla(uv)= \nabla(u)v+ u\nabla(v)[/itex] where u and v are scalar valued functions of variables x and y. This is, basically, just the "product rule" for gradients.

Okay,
[tex]\nabla(uv)= \frac{\partial uv}{\partial x}\vec{i}+ \frac{\partial uv}{\partial y}\vec{j}[/tex]
Apply the product rule to both partial derivatives:
[tex]\nabla(uv)= \left(u\frac{\partial v}{\partial x}+ \frac{\partial u}{\partial x}v\right)\vec{i}+ \left(u\frac{\partial v}{\partial y}+ \frac{\partial u}{\partial y}v\right)\vec{j}[/tex]
and regroup.

For part (a), if f(u, v)= 0 (where f is differentiable- that is NOT stated in the text but is necessary), then
[tex]\frac{\partial f}{\partial x}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}= 0[/tex]
and
[tex]\frac{\partial f}{\partial y}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}= 0[/tex]
Think of those as two equations to solve for the "unknowns" [itex]\partial f/\partial u[/itex] and [itex]\partial f/\partial v[/itex]. What condition on the "coefficients", [itex]\partial u/\partial x[/itex], [itex]\partial u/\partial y[/itex], [itex]\partial v/\partial x[/itex], and [itex]\partial v/\partial y[/itex] is necessary in order that that have a solution (think about the determinant of the matrix of coefficients- that's the point of (b)).
 
  • #4
Thank you very much, I think I've got it from here.
 

FAQ: Get Expert Tips for Proving the Product Rule Using Gradient Method

What is the product rule and why is it important in mathematics?

The product rule is a formula in calculus that is used to find the derivative of a product of two functions. It is important because it allows us to find the rate of change of a product, which is a common occurrence in many real-world problems.

What is the gradient method and how is it related to the product rule?

The gradient method is a technique used to find the maximum and minimum values of a multi-variable function. It is related to the product rule because it involves calculating the partial derivatives of a function, which is an essential step in using the product rule.

Can you explain how to use the gradient method to prove the product rule?

To prove the product rule using the gradient method, start by finding the partial derivatives of the product of two functions. Then, use the chain rule to expand each partial derivative. Finally, combine the expanded partial derivatives and simplify to show that it is equal to the derivative of the product using the product rule.

Are there any common mistakes to avoid when using the gradient method to prove the product rule?

One common mistake to avoid is not properly expanding the partial derivatives using the chain rule. This can result in an incorrect proof. It is also important to carefully keep track of all the steps and not make any calculation errors.

How can understanding the product rule and the gradient method benefit me as a scientist?

As a scientist, understanding the product rule and the gradient method can help you to accurately calculate rates of change and find maximum or minimum values in multi-variable functions. This is especially useful in fields such as physics, engineering, and economics.

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