Let $f$ be a differentiable function satisfying

$$f(x + y) = e^xf(y) + e^yf(x)$$

for all $x, y \in \mathbb{R}$. Find $f'(0)$.

I tried to use the definition of $f'(0)$ to do this:

$$f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0}\frac{f(h)}{h}$$

The problem here is that I would have to apply L'Hopital's rule on the RHS , which would give me

$$f'(0) = \frac{\lim_{h\to0}f'(h)}{\lim_{h\to0}1} = f'(0)$$

This doesn't really help much. Any ideas?


It is a trick question. A function satisfies that if $f(x) = cxe^x$ for any constant $c$. So $f'(0)=c$, and this can be any real number.

You can get the above functional form by: $\frac{f(x+h)-f(x)}{h} = \frac{e^xf(h)}{h} + \frac{f(x)(e^h-1)}{h}$
and so, taking limits and defining $\lim_{h\rightarrow 0} f(h)/h=c$ gives:
$f'(x) = ce^x + f(x)$.
So then we just solve the differential equation.


Given $f(x+y) = e^{y}f(x)+e^{x}f(y)$

Divide both side by $e^{x+y}\;,$ we get

$\displaystyle e^{-(x+y)}f(x+y) = e^{-x}f(x)+e^{-y}f(y)$

Now Let $\displaystyle e^{-x}f(x) = g(x)$, Then equation Convert into $g(x+y)=g(x)+g(y)$

Which is a Cauchy Functional equation whose solution is $g(x)=cx$.

So $\displaystyle g(x)=e^{-x}f(x)=cx\Rightarrow f(x)=cxe^{x}$

So $f{'}(x)=c\left\{xe^{x}+e^{x}\right\}$, So $f^{'}(0)=c$


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