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