Question

To find the derivative of the function.

Short answer

The derivative of the function \(y = {\mathop{\rm sech}\nolimits} \left( {{e^{ - x}}} \right)\) is, \(\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {e^{ - 2x}}} }}\).

Step-by-step solution

 Given function

The function is \(y = {\mathop{\rm sech}\nolimits} \left( {{e^{ - x}}} \right)\).

The Concept of derivative and hyperbolic function

Derivative rules:

(1) The chain rule:\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

(2) The derivative of the inverse hyperbolic sine function is,\(\frac{d}{{dx}}\left( {sin{h^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }}\).

Evaluate the derivative by the use of concepts

The derivative of the function \(y = {\mathop{\rm sech}\nolimits} \left( {{e^{ - x}}} \right)\) is computed as shown below.

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{\mathop{\rm sech}\nolimits} \left( {{e^{ - x}}} \right)} \right)\)

Let \(u = {e^{ - x}}\) and apply the chain rule.

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{d}{{dx}}({\mathop{\rm sech}\nolimits} u)\\ = \frac{d}{{du}}({\mathop{\rm sech}\nolimits} u)\frac{{du}}{{dx}}\\ = \frac{{ - 1}}{{u\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}\end{array}\)

Substitute, \(u = {e^{ - x}}\).

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{e^{ - x}}\sqrt {1 - {e^{ - 2x}}} }}\left( {\frac{d}{{dx}}\left( {{e^{ - x}}} \right)} \right)\\ = \frac{{ - 1}}{{{e^{ - x}}\sqrt {1 - {e^{ - 2x}}} }}\left( {{e^{ - x}}( - 1)} \right)\\ = \frac{{{e^{ - x}}}}{{{e^{ - x}}\sqrt {1 - {e^{ - 2x}}} }}\\ = \frac{1}{{\sqrt {1 - {e^{ - 2x}}} }}\end{array}\)

Therefore, the derivative of the function \(y = {\mathop{\rm sech}\nolimits} \left( {{e^{ - x}}} \right)\) is, \(\frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {e^{ - 2x}}} }}\).