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

• Jun 9th 2010, 04:20 PM
mathematicalbagpiper
Use The Chain Rule To Prove f+g is Differentiable
Let $\displaystyle f,g:E\longrightarrow F$ (finite normed vector spaces) be differentiable functions. Use the chain rule to prove that $\displaystyle f+g$ is also differentiable.

The chain rule has to do with composite functions, I know, so I know somehow I have to think of $\displaystyle f+g$ as a composite function, but I'm just not seeing it.
• Jun 19th 2010, 03:59 AM
Failure
Quote:

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

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

Let me see: You want to represent
$\displaystyle f+g\!:E\ni x\mapsto f(x)+g(x)\in F$
as the composition of two "functions", right?
$\displaystyle f\otimes g\!:E\ni x\mapsto (f(x),g(x))\in F^2$ and $\displaystyle +\!: F^2\ni (x,y) \mapsto x+y\in F$,
since you have that $\displaystyle f+g = +\circ \otimes$
(Surprised) Oops, that looks a bit funny, I must say. - What I mean is that $\displaystyle f+g$ equals "applying $\displaystyle +$ after $\displaystyle \otimes$", of course.
Remember that the map $\displaystyle (x,y)\mapsto x+y$ is differentiable, being a linear transformation.