Derivative of Hyperbolic Cotangent

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

Let the function of the form

y = f\left( x \right) = \coth x

By definition of hyperbolic function, the hyperbolic cotangent function is defined as

\coth x = \frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}

Now take 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 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}}}}, 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

Differentiation 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}