What is the proof for the property of taking the gradient of a function?

[ohlala191785]

Homework Statement

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

Operation: (∇(u))ⁿ = n*u^n-1*∇u

Homework Equations

The gradient vector of f is <∂f/∂x,∂f/∂y>, where f is a function of x and y (in other words f(x,y)).

The Attempt at a Solution

I tried proof by induction, but I have a lot of gaps.

Base, n=1: ∇u = n*u⁰∇u. But what is u⁰ if u is a function?

Assume n = k: ∇u^k = n*u^k-1∇u

For k=k+1, ∇u^k+1 = n*u^k∇u.

Isn't proof by induction for sums though? I can't seem to identify the sum here.Any help would be greatly appreciated!

 

[Dick]

Homework Statement

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

Operation: (∇(u))ⁿ = n*u^n-1*∇u

Homework Equations

The gradient vector of f is <∂f/∂x,∂f/∂y>, where f is a function of x and y (in other words f(x,y)).

The Attempt at a Solution

I tried proof by induction, but I have a lot of gaps.

Base, n=1: ∇u = n*u⁰∇u. But what is u⁰ if u is a function?

Assume n = k: ∇u^k = n*u^k-1∇u

For k=k+1, ∇u^k+1 = n*u^k∇u.

Isn't proof by induction for sums though? I can't seem to identify the sum here.


Any help would be greatly appreciated!

You have got to mean $\nabla (u^n)=n u^{n-1} \nabla u$, $(\nabla(u))^n$ doesn't mean anything. Just use the chain rule for partial derivatives.

 

[ohlala191785]

Oh I see. Thanks!