- #1
cliowa
- 191
- 0
Let [tex]L(A;B)[/tex] be the space of linear maps [tex]l:A\rightarrow B[/tex].
My goal is to derive the Leibniz (Product) Rule using the chain rule. Let [tex]f_i:U\subset E\rightarrow F_i, i=1,2[/tex] be differentiable maps and let [tex]B\in L(F_1,F_2;G)[/tex]. Then the mapping [tex]B(f_1,f_2)=B\circ (f_1\times f_2):U\subset E\rightarrow G[/tex] is differentiable.
Using the chain rule I find for [tex]u\in U[/tex]: [tex]D \{B\circ (f_1\times f_2)\}(u)=DB \{(f_1\times f_2)(u)\}\circ D(f_1\times f_2)(u)=B\circ D(f_1\times f_2)(u)\in L(E;G)[/tex], as B is a linear map. I could now go on and take the derivative of the cartesian product: [tex]...=B\circ(Df_1\times Df_2)(u)[/tex], but I feel it's all wrong. To me, there's no product rule in sight. Could you point out my error(s)?
Thanks alot. Best regards...Cliowa
My goal is to derive the Leibniz (Product) Rule using the chain rule. Let [tex]f_i:U\subset E\rightarrow F_i, i=1,2[/tex] be differentiable maps and let [tex]B\in L(F_1,F_2;G)[/tex]. Then the mapping [tex]B(f_1,f_2)=B\circ (f_1\times f_2):U\subset E\rightarrow G[/tex] is differentiable.
Using the chain rule I find for [tex]u\in U[/tex]: [tex]D \{B\circ (f_1\times f_2)\}(u)=DB \{(f_1\times f_2)(u)\}\circ D(f_1\times f_2)(u)=B\circ D(f_1\times f_2)(u)\in L(E;G)[/tex], as B is a linear map. I could now go on and take the derivative of the cartesian product: [tex]...=B\circ(Df_1\times Df_2)(u)[/tex], but I feel it's all wrong. To me, there's no product rule in sight. Could you point out my error(s)?
Thanks alot. Best regards...Cliowa
Last edited: