# Use The Chain Rule To Prove f+g is Differentiable

• Jun 9th 2010, 05:20 PM
mathematicalbagpiper
Use The Chain Rule To Prove f+g is Differentiable
Let $f,g:E\longrightarrow F$ (finite normed vector spaces) be differentiable functions. Use the chain rule to prove that $f+g$ is also differentiable.

The chain rule has to do with composite functions, I know, so I know somehow I have to think of $f+g$ as a composite function, but I'm just not seeing it.
• Jun 19th 2010, 04:59 AM
Failure
Quote:

Originally Posted by mathematicalbagpiper
Let $f,g:E\longrightarrow F$ (finite normed vector spaces) be differentiable functions. Use the chain rule to prove that $f+g$ is also differentiable.

The chain rule has to do with composite functions, I know, so I know somehow I have to think of $f+g$ as a composite function, but I'm just not seeing it.

Let me see: You want to represent
$f+g\!:E\ni x\mapsto f(x)+g(x)\in F$
as the composition of two "functions", right?
$f\otimes g\!:E\ni x\mapsto (f(x),g(x))\in F^2$ and $+\!: F^2\ni (x,y) \mapsto x+y\in F$,
since you have that $f+g = +\circ \otimes$
(Surprised) Oops, that looks a bit funny, I must say. - What I mean is that $f+g$ equals "applying $+$ after $\otimes$", of course.
• Jun 19th 2010, 08:21 AM
Bruno J.
Failure has it right! (Funny thing to say (Rofl))

Remember that the map $(x,y)\mapsto x+y$ is differentiable, being a linear transformation.