Derivative of Hyperbolic Secant

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

Let the function of the form

y = f\left( x \right) = \operatorname{sech} x

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

\operatorname{sech} x = \frac{2}{{{e^x} + {e^{ - x}}}}

Differentiating both sides with respect to the variable x, we have

\begin{gathered} \frac{d}{{dx}}\operatorname{sech} x = \frac{d}{{dx}}\left( {\frac{2}{{{e^x} + {e^{ - x}}}}} \right) \\ \Rightarrow \frac{d}{{dx}}\operatorname{sech} x = 2\frac{d}{{dx}}{\left( {{e^x} + {e^{ - x}}} \right)^{ - 1}} \\ \Rightarrow \frac{d}{{dx}}\operatorname{sech} x = - 2{\left( {{e^x} + {e^{ - x}}} \right)^{ - 2}}\frac{d}{{dx}}\left( {{e^x} + {e^{ - x}}} \right) \\ \end{gathered}

Using the formula of exponential differentiation \frac{d}{{dx}}{e^x} = {e^x}, we have

\begin{gathered} \frac{d}{{dx}}\operatorname{sech} x = - 2{\left( {{e^x} + {e^{ - x}}} \right)^{ - 2}}\left( {{e^x} - {e^{ - x}}} \right) \\ \Rightarrow \frac{d}{{dx}}\operatorname{sech} x = - 2\frac{{\left( {{e^x} - {e^{ - x}}} \right)}}{{{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}} \\ \Rightarrow \frac{d}{{dx}}\operatorname{sech} x = - \frac{2}{{{e^x} + {e^{ - x}}}}\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \\ \end{gathered}

By definition, \operatorname{sech} x = \frac{2}{{{e^x} + {e^{ - x}}}} and \tanh x = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}, we get

\frac{d}{{dx}}\operatorname{sech} x = - \sec hx\tanh x

Example: Find the derivative of

y = f\left( x \right) = \operatorname{sech} {x^3}

We have the given function as

y = \operatorname{sech} {x^3}

Differentiation with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}\operatorname{sech} {x^3}

Using the rule, \frac{d}{{dx}}\operatorname{sech} x = - \sec hx\tanh x, we get

\begin{gathered} \frac{{dy}}{{dx}} = - \operatorname{sech} {x^3}\tanh {x^3}\frac{d}{{dx}}{x^3} \\ \Rightarrow \frac{{dy}}{{dx}} = - \operatorname{sech} {x^3}\tanh {x^3}\left( {3{x^2}} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = - 3{x^2}\operatorname{sech} {x^3}\tanh {x^3} \\ \end{gathered}