Answer
\[{y^,} = \frac{{x{e^x} + x{e^{ - x}} - {e^x} + {e^{ - x}}}}{{{x^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{{e^x} - {e^{ - x}}}}{x} \hfill \\
Differentiate\,\,using\,\,the\,\,quotient\,\,rule \hfill \\
\,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{x\,{{\left( {{e^x} - {e^{ - x}}} \right)}^,}\,\left( {{e^x} - {e^{ - x}}} \right)\,{{\left( x \right)}^,}}}{{\,{{\left( x \right)}^2}}} \hfill \\
Compute\,\,the\,\,derivatives \hfill \\
{y^,} = \frac{{x\,\left( {{e^x} + {e^{ - x}}} \right) - \,\left( {{e^x} - {e^{ - x}}} \right)\,\left( 1 \right)}}{{{x^2}}} \hfill \\
Multiply \hfill \\
{y^,} = \frac{{x{e^x} + x{e^{ - x}} - {e^x} + {e^{ - x}}}}{{{x^2}}} \hfill \\
\end{gathered} \]