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}

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