# Derivative of Hyperbolic Secant

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

Let the function be of the form $y = f\left( x \right) = \operatorname{sech} x$

By the definition of the 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 $\tanh x = \frac{{{e^x} – {e^{ – x}}}}{{{e^x} + {e^{ – x}}}}$, so 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
$y = \operatorname{sech} {x^3}$

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