Gradient vector property proofs

[fastXattack]

Homework Statement

Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differentiable functions of x and y and that a, b are constants.

Homework Equations

Δ = gradient vector

1) Δ(u/v) = vΔu - uΔv / v^2

2) Δu^n = nu^(n-1)Δu

The Attempt at a Solution

I tried taking the partial derivative of (u/v) and separating the terms but I didn't get the ending result.

For 2, I don't even know where to begin because it is an exponent.

 

[Mark44]

Homework Statement

Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differentiable functions of x and y and that a, b are constants.

Homework Equations

Δ = gradient vector

1) Δ(u/v) = vΔu - uΔv / v^2

2) Δu^n = nu^(n-1)Δu

The Attempt at a Solution

I tried taking the partial derivative of (u/v) and separating the terms but I didn't get the ending result.

For 2, I don't even know where to begin because it is an exponent.

Let's try a simpler problem:
$$\nabla(uv) = v\nabla{u} + u\nabla{v}$$

$$\nabla(uv) = <\frac{\partial (uv)}{\partial x}, \frac{\partial (uv)}{\partial y}>$$
$$= <u\frac{\partial v}{\partial x} + v\frac{\partial u}{\partial x}, u\frac{\partial v}{\partial y} + v\frac{\partial u}{\partial y}>$$
$$= u< \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}> + v< \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} > = u\nabla{v} + v\nabla{u}$$

 

[fastXattack]

I was given that problem for homework as well and I was able to do that one. I have an idea on how to do the quotient rule one, but no idea for the second question.

 

[HallsofIvy]

If (2) is to be done for n a positive integer, use induction on n.

 

[fastXattack]

This is my attempt at the quotient rule proof. Did I do the correct thing for this problem at least? I'm still unsure how to do the second question...

 

[Mark44]

Yes, that looks fine although the second line is difficult to read.

For the other problem, use induction (the tacit assumption is that n is a positive integer). The base case is easy to establish.
$$\nabla u^1 = \nabla u = 1 \nabla u$$

Now assume that for n = k,
$$\nabla u^k = k \nabla u^{k - 1}$$

To complete the proof, show that for n = k + 1
$$\nabla u^{k + 1} = (k + 1) \nabla u^k$$