\displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\tanh(x + h) - \tanh x}{h} = \lim_{h \to 0} \frac{\frac{e^{x+h} - e^{-x-h}}{e^{x+h} + e^{-x -h}} - \frac{e^x - e^{-x}}{e^x + e^{-x}}}{h} = \lim_{h \to 0} \frac{\frac{e^{2x+2h} - 1}{e^{2x+2h} + 1} - \frac{e^{2x} - 1}{e^{2x}+1}}{h} = \lim_{h \to 0} \frac{\frac{e^{2x}e^{2h} - 1}{e^{2x}e^{2h} + 1} - \frac{e^{2x} - 1}{e^{2x}+1}}{h} = \lim_{h \to 0} \frac{(e^{2x}e^{2h} - 1)(e^{2x}+1)-(e^{2x} - 1)(e^{2x}e^{2h} + 1)}{h(e^{2x}e^{2h} + 1)(e^{2x}+1)} = \lim_{h \to 0} \frac{2 e^{2x} e^{2h} - 2 e^{2x}}{h(e^{2x}e^{2h} + 1)(e^{2x}+1)} = \lim_{h \to 0} \frac{2e^{2x}(e^{2h}-1)}{h(e^{2x}e^{2h} + 1)(e^{2x}+1)} = \lim_{h \to 0} \frac{e^{2h}-1}{2h} \cdot \frac{4e^{2x}}{(e^{2x}e^{2h} + 1)(e^{2x}+1)} = \left( \frac{2 e^x}{e^{2x}+1} \right)^2 = \frac{1}{\cosh^2 x} |