1. ## Exponential Function

On Page 78 of Lang - Undergraduate Analysis (2nd Ed) we find the following:
"We assume that there is a function f defined for all numbers such that f ' = f
and f(0) = 1.
We note that f(x)is not equal to 0 for all x. Indeed differentiating the function
f(x)f(-x) we find 0"

Can anyone help me show rigorously that the derivative of f(x)f(-x) for the exponential function is zero - relying only on f ' = f and f(0) = 1.

Help will be appreciated.

2. Originally Posted by Bernhard
On Page 78 of Lang - Undergraduate Analysis (2nd Ed) we find the following:
"We assume that there is a function f defined for all numbers such that f ' = f
and f(0) = 1.
We note that f(x)is not equal to 0 for all x. Indeed differentiating the function
f(x)f(-x) we find 0"

Can anyone help me show rigorously that the derivative of f(x)f(-x) for the exponential function is zero - relying only on f ' = f and f(0) = 1.

Help will be appreciated.
Use the product rule: the derivative of f(x)f(-x) is f'(x)f(-x)+ f(x)(f(-x))' which, by the chain rule is f'(x)f(-x)- f(x)f'(-x). Since we have that f'(x)= f(x), we also have that f'(-x)= f(-x) and so (f(x)f(-x))'= f(x)f(-x)- f(x)f(-x)= 0.

(You don't need f(0)= 1. This is true for any exponential $Ae^x$. $(Ae^x)(Ae^{-x})= A^2$ which is a constant and has derivative 0.)

3. ## Thanks HallsofIvy. appreciate your help! Can you help further?

Thanks.

I think my problem was how to differentiate f(-x) but I should have used chain rule for composite functions, so

d/dx [f(-x)] = d/dx f(-x) . d/dx (-x)

= d/dx f(x) . -1

= - f ' (x)

Then when you put this into the product rule the result follows

BUT

I am a bit uncomfortable with the formal step d/dx f(-x) = d/dx f(x)

Can you explain how this follows exactly

4. Originally Posted by Bernhard
Thanks.

Can you explain how you show rigorously that f ' (-x) = f(-x). Why, exactly, does this follow?
$f'(x) = f(x) \ \forall x \in \mathbb{R}.$
Let $t = -x \Rightarrow f'(t) = f(t)$ by the above assumption, and thus $f'(-x) = f(-x)$