Answer
$$\frac{d}{{dx}}\left[ {\coth x} \right] = - {\operatorname{csch} ^2}x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\coth x} \right] \cr
& {\text{Using hyperbolic identities}} \cr
& \frac{d}{{dx}}\left[ {\frac{{\cosh x}}{{\sinh x}}} \right] \cr
& {\text{By the definition of }}\cosh x{\text{ and sinh }}x \cr
& \frac{d}{{dx}}\left[ {\frac{{\cosh x}}{{\sinh x}}} \right] = \frac{d}{{dx}}\left[ {\frac{{\frac{{{e^x} + {e^{ - x}}}}{2}}}{{\frac{{{e^x} - {e^{ - x}}}}{2}}}} \right] = \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] \cr
& {\text{Differentiating and simplifying}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] = \frac{{\left( {{e^x} - {e^{ - x}}} \right)\left( {{e^x} - {e^{ - x}}} \right) - \left( {{e^x} + {e^{ - x}}} \right)\left( {{e^x} + {e^{ - x}}} \right)}}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] = \frac{{{e^{2x}} - 2 + {e^{ - 2x}} - {e^{2x}} - 2 - {e^{ - 2x}}}}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] = \frac{{ - 4}}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] = - {\left( {\frac{2}{{{e^x} - {e^{ - x}}}}} \right)^2} \cr
& {\text{Where }}\operatorname{csch} x = \frac{2}{{{e^x} - {e^{ - x}}}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}} \right] = - {\operatorname{csch} ^2}x \cr
& {\text{Proved}}{\text{.}} \cr
& \frac{d}{{dx}}\left[ {\coth x} \right] = - {\operatorname{csch} ^2}x \cr} $$
