Math Doubts

Derivative of Hyperbolic Secant function


$\dfrac{d}{dx}{\, \operatorname{sech}{x}}$ $\,=\,$ $-\operatorname{sech}{x}\tanh{x}$


When $x$ represents a variable, the hyperbolic secant function is written as $\operatorname{sech}{x}$ in mathematical form. The derivative of the hyperbolic secant function with respect to $x$ is written in below form in differential calculus.

$\dfrac{d}{dx}{\, \operatorname{sech}{x}}$

The differentiation formula of the hyperbolic secant function is simply written mathematically as $(\operatorname{sech}{x})’$ in calculus. The differentiation of the hyperbolic secant function is equal to the negative sign of product of hyperbolic secant and tangent functions.

$\implies$ $\dfrac{d}{dx}{\, \operatorname{sech}{x}}$ $\,=\,$ $-\operatorname{sech}{x}\tanh{x}$

Other forms

Mathematically, the derivative of hyperbolic secant function can also be written in terms of any variable in differential calculus.


$(1) \,\,\,$ $\dfrac{d}{dl}{\, \operatorname{sech}{(l)}}$ $\,=\,$ $-\operatorname{sech}{(l)}\tanh{(l)}$

$(2) \,\,\,$ $\dfrac{d}{dq}{\, \operatorname{sech}{(q)}}$ $\,=\,$ $-\operatorname{sech}{(q)}\tanh{(q)}$

$(3) \,\,\,$ $\dfrac{d}{dy}{\, \operatorname{sech}{(y)}}$ $\,=\,$ $-\operatorname{sech}{(y)}\tanh{(y)}$


Learn how to prove the differentiation of hyperbolic secant in differential calculus from the first principle of differentiation.

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