# Effortlessly Solve C^{2} Functions with Quick and Simple Tips | Get Help Now!

• coolalac
In summary, the problem is to show that for a C^{2} function u in R^{n} and an nxn matrix A, the gradient of the function v(x) defined as v(x):= u(Ax) is equal to A^{T} . \nabla u(Ax). This can be solved using the chain rule.
coolalac

Hey I was wondering if you'll could help. I was told that this is a simple proof but have been working on it for ages and can't get it:

Suppose u is a $$C^{2}$$ function in $$R^{n}$$. Given an nxn matrix, A define v(x):= u(Ax). Show that ($$\nabla$$v)(x) = $$A^{T}$$ . $$\nabla$$u(Ax)

(Note the grad u(Ax) isn't supposed to be superscripted I just don't know how to put it in the same line!)

Thanks

coolalac said:
Hey I was wondering if you'll could help. I was told that this is a simple proof but have been working on it for ages and can't get it:

Suppose u is a $$C^{2}$$ function in $$R^{n}$$. Given an nxn matrix, A define v(x):= u(Ax). Show that ($$\nabla$$v)(x) = $$A^{T}$$ . $$\nabla$$u(Ax)

(Note the grad u(Ax) isn't supposed to be superscripted I just don't know how to put it in the same line!)
You put every thing on the same line by putting the entire formula inside [ tex ] and [ /tex ], not just bits and pieces:
$$\nabla v(x)= A^T\nambla u(Ax)$$
Much simpler to write and to read!

Thanks
It is, essentially, the "chain rule". v(x)= u(Ax) so $\nabla v(x)= \nabla u(x) d(Ax)/dt= \nabla u(x) A = A^T \nabla u(x)$.

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