Two questions I suppose. One comes from a test I recently took that I didn't quite get/understand the method I should be using (or even how I should proceed)
Let $f:R^2 \rightarrow R$ s.t $f$ is an element of $C^2$(i.e. continuous first/second partials). Let $h(s,t)=(s+t,s-t)$.
a) Show that $F=f∘h$ is differentiable and its first partials are also differentiable. [Note - no problem here. I got this problem correct. Just putting it here for context]
b) Prove that $D_{1,2}(f∘h)=D_{1,1}(f∘h)-D_{2,2}(f∘h)$
On b I guess I'm just not sure how to proceed.
What I tried to do via playing around with it(though I think I'm wrong about my attempts to think about it):
$f(h(s,t)=f(s+t,s-t)$
$D_1(f∘h)= D_1f*Dh_1+D_2f*D_1h_2 = D_1f+D_2f$
$D_{1,1}(f∘h)=...$
I wrote something for this but it was pointed out that it had to be wrong...I'm not sure if I can evaluate $D_1f+D_2f$ further but someone made me think so. So atm, I'm just confused on how to proceed. [I decided to make something separate for the other question so nevermind on the second question].
EDIT: With a prompt elsewhere, I may have found a way forward that I wasn't considering. The test question asked me for $D_{1,2}(f∘h)=D_{1,1}(f∘h)-D_{2,2}(f∘h)$
My initial thought was to calculate $D_{1,1}(f∘h)-D_{2,2}(f∘h)$ and set it equal to the LHS to tease the proof out by some difference. Instead I went ahead with a direct route:
$D_{2}F=D_{1}fD_{2}h+D_{2}fD_{2}=D_{1}f-D_{2}f$
Then:
$D_{1,2}F=(D_{1,1}f-D_{2,1}h)+(D_{2,1}f-D_{2,2}f$)
Since F is an element of C^2(R^2) => $D_{1,2}f=D_{2,1}f$
Then $D_{1,2}F=D_{1,1}-D_{2,2}f$
which implies:
$D_{1,2}(f∘h)=D_{1,1}(f∘h)-D_{2,2}(f∘h)$
So, I may or may not have it figured out... depending if there is a mistake in my logic.