Derivative of Hyperbolic Cotangent

In this tutorial we shall prove the derivative of the hyperbolic cotangent function.

Let the function be of the form \[y = f\left( x \right) = \coth x\]

By the definition of the hyperbolic function, the hyperbolic cotangent function is defined as
\[\coth x = \frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\]

Now taking this function for differentiation, we have
\[\coth x = \frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}\]

Differentiating both sides with respect to the variable $$x$$, we have
\[\frac{d}{{dx}}\coth x = \frac{d}{{dx}}\left( {\frac{{{e^x} + {e^{ – x}}}}{{{e^x} – {e^{ – x}}}}} \right)\]

Using the quotient formula for differentiation, we have
\[\frac{d}{{dx}}\left( {\coth x} \right) = \frac{{\left( {{e^x} – {e^{ – x}}} \right)\frac{d}{{dx}}\left( {{e^x} + {e^{ – x}}} \right) – \left( {{e^x} + {e^{ – x}}} \right)\frac{d}{{dx}}\left( {{e^x} – {e^{ – x}}} \right)}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}}\]

Using the formula of exponential differentiation $$\frac{d}{{dx}}{e^x} = {e^x}$$, we have
\[\begin{gathered} \frac{d}{{dx}}\left( {\coth x} \right) = \frac{{\left( {{e^x} – {e^{ – x}}} \right)\left( {{e^x} – {e^{ – x}}} \right) – \left( {{e^x} + {e^{ – x}}} \right)\left( {{e^x} + {e^{ – x}}} \right)}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = \frac{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2} – {{\left( {{e^x} + {e^{ – x}}} \right)}^2}}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = \frac{{\left( {{e^{2x}} + {e^{ – 2x}} – 2{e^x}{e^{ – x}}} \right) – \left( {{e^{2x}} + {e^{ – 2x}} + 2{e^x}{e^{ – x}}} \right)}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = \frac{{\left( {{e^{2x}} + {e^{ – 2x}} – 2} \right) – \left( {{e^{2x}} + {e^{ – 2x}} + 2} \right)}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = \frac{{{e^{2x}} + {e^{ – 2x}} – 2 – {e^{2x}} – {e^{ – 2x}} – 2}}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = – \frac{4}{{{{\left( {{e^x} – {e^{ – x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = – {\left( {\frac{2}{{{e^x} – {e^{ – x}}}}} \right)^2} \\ \end{gathered} \]

By definition, $$\operatorname{csch} x = \frac{2}{{{e^x} – {e^{ – x}}}}$$, so we get
\[ \Rightarrow \frac{d}{{dx}}\left( {\coth x} \right) = – \csc {h^2}x\]

 

Example: Find the derivative of \[y = f\left( x \right) = \coth 4x\]

We have the given function as
\[y = \coth 4x\]

Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}\coth 4x\]

Using the rule, $$\frac{d}{{dx}}\left( {\coth x} \right) = – \csc {h^2}x$$, we get
\[\begin{gathered} \frac{{dy}}{{dx}} = – {\operatorname{csch} ^2}4x\frac{d}{{dx}}4x \\ \Rightarrow \frac{{dy}}{{dx}} = – {\operatorname{csch} ^2}4x\left( 4 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = – 4{\operatorname{csch} ^2}4x \\ \end{gathered} \]