The definition of the derivative \(f'\) of a (differentiable) function f is
$$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}. \quad (*)$$
The function f has derivative \(f'\) and satisfies
$$f(x + y) = f(x)f(y)$$
for all \(x\) and \(y\), and \(f'(0) = k\) where \(k \neq 0\). Show that \(f(0) = 1\).
Using \((*)\), show that \(f'(x) = kf(x)\) and find \(f(x)\) in terms of \(x\) and \(k\).
The function g has derivative \(g'\) and satisfies
$$g(x + y) = \frac{g(x) + g(y)}{1 + g(x)g(y)}$$
for all \(x\) and \(y\), \(|g(x)| < 1\) for all \(x\), and \(g'(0) = k\) where \(k \neq 0\).
Find \(g'(x)\) in terms of \(g(x)\) and \(k\), and hence find \(g(x)\) in terms of \(x\) and \(k\).
Solution:
\(\,\) \begin{align*}
&& f(0+x) &= f(0)f(x) \\
\Rightarrow && f(0) &= 0, 1\\
&&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\
\Rightarrow && f(0) &= 1
\end{align*}
\begin{align*}
&& f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\
&&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\
&&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\
&&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\
&&&= f(x) \cdot f'(0) \\
&&&= kf(x)
\end{align*}
Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)