abstract function differentiation?

Question

G(A,B) = F(A + B, A - B)

Where f is a differentiable function of two variables, F = F(U, V)

Show that:
dG dG
dA dB

Is equal to

dF dF Minus dF dF
dU dU _____ dV dV

It's difficult to write fractions. Basically, I need help showing that the second derivative of G, with respect to A and B, is equal to the derivative of F with respect to U squared minus the derivative of F with respect to V squared.

This one is killing me, I can't sleep right anymore!

Answer 1

G(A, B) = F(A+B, A-B)

By the chain rule, ∂G/∂A = ∂F/∂U * ∂U/∂A + ∂F/∂V * ∂V/∂A. Since in this case we set U=A+B and V=A-B, we have ∂U/∂A = 1 and ∂V/∂A = 1, so ∂G/∂A = ∂F/∂U + ∂F/∂V.

Now, taking the derivative again:

∂²G/(∂A∂B) = ∂(∂F/∂U)/∂B + ∂(∂F/∂V)/∂B

Employing the chain rule again on both functions:

∂²G/(∂A∂B) = ∂²F/∂U² * ∂U/∂B + ∂²F/(∂U∂V) * ∂V/∂B + ∂²F/(∂V∂U) * ∂U/∂B + ∂²F/∂V² * ∂V/∂B

Now ∂U/∂B = 1 and ∂V/∂B = -1, so this becomes:

∂²G/(∂A∂B) = ∂²F/∂U² - ∂²F/(∂U∂V) + ∂²F/(∂V∂U) - ∂²F/∂V²

Now, if we assume that the second partial derivatives at this point are continuous, then by Clairaut's theorem, ∂²F/(∂U∂V) = ∂²F/(∂V∂U) and so the middle terms cancel out, leaving us with:

∂²G/(∂A∂B) = ∂²F/∂U² - ∂²F/∂V²

As required.