## Calculus, 10th Edition (Anton)

$$y' = \frac{4}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}$$
\eqalign{ & y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr & {\text{use quotient rule }} \cr & \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr & y' = \frac{{\left( {{e^x} + {e^{ - x}}} \right)\left[ {{e^x} - {e^{ - x}}} \right]' - \left( {{e^x} - {e^{ - x}}} \right)\left[ {{e^x} + {e^{ - x}}} \right]'}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr & y' = \frac{{\left( {{e^x} + {e^{ - x}}} \right)\left( {{e^x} + {e^{ - x}}} \right) - \left( {{e^x} - {e^{ - x}}} \right)\left( {{e^x} - {e^{ - x}}} \right)}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr & {\text{multiply}} \cr & y' = \frac{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2} - {{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr & {\text{simplify}} \cr & y' = \frac{{{e^{2x}} + 2{e^{x - x}} + {e^{ - 2x}} - {e^{2x}} + 2{e^{x - x}} - {e^{ - 2x}}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr & y' = \frac{{{e^{2x}} + 2 + {e^{ - 2x}} - {e^{2x}} + 2 - {e^{ - 2x}}}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr & y' = \frac{4}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \cr}