\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let \(y = 0\), and we have \[ f(x + 0) = f(x) = f(x) f(0), \] so either \(f(x) = 0\) or \(f(0) = 1\) for all \(x\).
Assume, B.W.O.C., that \(f(0) \neq 1\), then we must have \(f(x) = 0\) for all \(x\), which means \(f'(x) = 0\), contradicting with \(f'(0) = k \neq 0\).
Hence, \(f(0) = 1\).
By definition of the derivative, we have \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) \lim _{h \to 0} \frac {f(h) - 1}{h}, \end {align*}
and letting \(x = 0\), we also have \[ k = f'(0) = f(0) \lim _{h \to 0} \frac {f(h) - 1}{h} = \lim _{h \to 0} \frac {f(h) - 1}{h}, \] and hence \[ f'(x) = k f(x) \] as desired.
This differential equation solves to \[ f(x) = A e^{kx}, \] and with the condition \(f(0) = 1\), we have \(A = 1\), and hence \[ f(x) = e^{kx} \] for all \(x\).
Let \(y = 0\), and we have \[ g(x + 0) = g(x) = \frac {g(x) + g(0)}{1 + g(x) g(0)}. \]
This means that \[ g(x) + g(x)^2 g(0) = g(x) + g(0), \] which gives \[ g(0) \left [g(x)^2 - 1\right ] = 0. \]
Since \(\abs *{g(x)} < 1\) for all \(x\), we must have \(g(x)^2 - 1 < 0\), and hence \(g(0) = 0\).
By the definition of the derivative, \begin {align*} g'(x) & = \lim _{h \to 0} \frac {g(x + h) - g(x)}{h} \\ & = \lim _{h \to 0} \frac {\frac {g(x) + g(h)}{1 + g(x) g(h)} - g(x)}{h} \\ & = \lim _{h \to 0} \frac {g(x) + g(h) - g(x) - g(x)^2 g(h)}{h (1 + g(x) g(h))} \\ & = \lim _{h \to 0} \frac {g(h) \left [1 - g(x)^2\right ]}{h (1 + g(x) g(h))} \\ & = \left [1 - g(x)^2\right ] \lim _{h \to 0} \frac {g(h)}{h (1 + g(x) g(h))}. \end {align*}
Considering the limit, we have \begin {align*} \lim _{h \to 0} \frac {g(h)}{h (1 + g(x) g(h))} & = \lim _{h \to 0} \frac {g(h) / h}{1 + g(x) g(h)} \\ & = \frac {\lim _{h \to 0} \left [g(h) / h\right ]}{\lim _{h \to 0} \left [1 + g(x) g(h)\right ]} \\ & = \frac {\lim _{h \to 0} \left [g(h) / h\right ]}{1} \\ & = \lim _{h \to 0} \frac {g(h)}{h}, \end {align*}
and hence \[ g'(x) = \left [1 - g(x)^2\right ] \lim _{h \to 0} \frac {g(h)}{h}. \]
Let \(x = 0\), and we have \[ k = g'(0) = 1 \cdot \lim _{h \to 0} \frac {g(h)}{h}, \] hence giving the differential equation \[ g'(x) = k \left [1 - g(x)^2\right ]. \]
This rearranges to give \[ \frac {\Diff g(x)}{1 - g(x)^2} = k \Diff x, \] and hence \[ \left [\frac {1}{1 + g(x)} + \frac {1}{1 - g(x)}\right ] \Diff g(x) = 2 k \Diff x, \] which gives \[ \ln \abs *{1 + g(x)} - \ln \abs *{1 - g(x)} = 2kx + C. \]
Let \(x = 0\), we have \(g(0) = 0\), and hence \(C = 0\), and hence \[ \frac {1 + g(x)}{1 - g(x)} = \exp (2kx), \] and hence \[ 1 + g(x) = \exp (2kx) - \exp (2kx) g(x), \] which gives \[ g(x) = \frac {\exp (2kx) - 1}{\exp (2kx) + 1} = \frac {\exp (kx) - \exp (-kx)}{\exp (kx) + \exp (-kx)} = \tanh (kx). \]