Math Doubts

Proof of Derivative of Hyperbolic Secant function

The hyperbolic secant function is mathematically written as $\operatorname{sech}{x}$ when $x$ is used to represent a variable. The differentiation or the derivative of hyperbolic secant function with respect to $x$ is written in the below mathematical form.

$\dfrac{d}{dx}{\, \Big(\operatorname{sech}{(x)}\Big)}$

The derivative of the hyperbolic secant function is proved by the first principle of the differentiation in differential calculus. Therefore, let us learn how to derive the differentiation formula for the hyperbolic secant function.

Derivative of Hyperbolic Secant function in Limit form

The derivative of hyperbolic secant function is mainly derived in limit form from the fundamental definition of the derivative in differential calculus.

$\dfrac{d}{dx}{\, (\operatorname{sech}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\operatorname{sech}{(x+\Delta x)}-\operatorname{sech}{x}}{\Delta x}}$

Let $\Delta x$ is represented by $h$ simply, then it is written in terms of $h$ instead of $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, (\operatorname{sech}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\operatorname{sech}{(x+h)}-\operatorname{sech}{x}}{h}}$

The derivative of $\operatorname{sech}{(x)}$ function with respect to $x$ is proved in differential calculus by the first principle of differentiation.

Evaluate the Limit by the Direct Substitution

Let’s use the direct substitution method to evaluate the differentiation of hyperbolic secant function as $h$ approaches zero.

$= \,\,\,$ $\dfrac{\operatorname{sech}{(x+0)}-\operatorname{sech}{x}}{0}$

$= \,\,\,$ $\dfrac{\operatorname{sech}{x}-\operatorname{sech}{x}}{0}$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\operatorname{sech}{x}}-\cancel{\operatorname{sech}{x}}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

The differentiation of hyperbolic secant function should not be equal to indeterminate. Hence, the direct substitution method is not recommendable in this case. Therefore, we must use an alternative method for proving the derivative of $\operatorname{sech}{x}$ function.

Simplify the Hyperbolic Secant function

Let us concentrate on the definition for the differentiation of secant function in limit form.

$\implies$ $\dfrac{d}{dx}{\, (\operatorname{sech}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\operatorname{sech}{(x+h)}-\operatorname{sech}{x}}{h}}$

It can be simplified by expressing every hyperbolic secant function in terms of natural exponential functions.

$\implies$ $\dfrac{d}{dx}{\, (\operatorname{sech}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{2}{e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}} -\dfrac{2}{e^{\displaystyle x}+e^{\displaystyle -x}} }{h}}$

Now, we have to concentrate on simplifying the expression in the right hand side of the equation.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{ \dfrac{2\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) -2\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big) }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) } }{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) -2\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big) }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) -\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[e^{\displaystyle x}+e^{\displaystyle -x}-e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[e^{\displaystyle x}+e^{\displaystyle -x}-e^{\displaystyle x+h}-e^{\displaystyle -x-h}\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

The third and fourth terms in the second factor of the numerator can be split by the product rule of exponents.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[e^{\displaystyle x}+e^{\displaystyle -x}-e^{\displaystyle x} \times e^{\displaystyle h}-e^{\displaystyle -x} \times e^{\displaystyle -h}\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[e^{\displaystyle x}-e^{\displaystyle x} \times e^{\displaystyle h}+e^{\displaystyle -x}-e^{\displaystyle -x} \times e^{\displaystyle -h}\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2\Bigg[e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)\Bigg] }{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times h}}$

The function can be split as product of two functions.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{2}{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}}$ $\times$ $\dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}\Bigg)$

Now, let us evaluate the limit of the product of the functions by the product rule of limits.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{2}{\Big(e^{\displaystyle (x+h)}+e^{\displaystyle -(x+h)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}}$

Evaluate the Limit of the function

The limit of the function can be evaluated by evaluating the product of the limits of the functions. Firstly, let us evaluate the limit of the first multiplying function by the direct substitution method.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle (x+0)}+e^{\displaystyle -(x+0)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle (x)}+e^{\displaystyle -(x)}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}}$

It is time to evaluate the second factor but the direct substitution method is not recommendable in this case. So, let us try to evaluate it in alternative mathematical approach.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(1 \times \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{(-1)}{(-1)} \times \dfrac{e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(-1) \times \Bigg(e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)\Bigg)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(-1) \times e^{\displaystyle x}\Big(1-e^{\displaystyle h}\Big)+(-1) \times e^{\displaystyle -x}\Big(1-e^{\displaystyle -h}\Big)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(-1) \times e^{\displaystyle x} \times \Big(1-e^{\displaystyle h}\Big)+(-1) \times e^{\displaystyle -x} \times \Big(1-e^{\displaystyle -h}\Big)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x} \times (-1) \times \Big(1-e^{\displaystyle h}\Big)+e^{\displaystyle -x} \times (-1) \times \Big(1-e^{\displaystyle -h}\Big)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x} \times \Big(-1+e^{\displaystyle h}\Big)+e^{\displaystyle -x} \times \Big(-1+e^{\displaystyle -h}\Big)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)+e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{(-1) \times h}}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)+e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{(-1) \times h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{(-1) \times h}+\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{(-1) \times h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{1 \times e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{(-1) \times h}+\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{(-1)} \times \dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{h}+\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg((-1) \times \dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{h}+\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(-\dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{h}+\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}\Bigg)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}-\dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{h}\Bigg)}$

The limit of the difference of the functions can be evaluated by the difference of their limits. Hence, use difference rule of the limits.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -x} \times \Big(e^{\displaystyle -h}-1\Big)}{-h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x} \times \Big(e^{\displaystyle h}-1\Big)}{h}\Bigg)}$

In this expression, the natural exponential functions $e^{\displaystyle -x}$ and $e^{\displaystyle x}$ are constants. Therefore, they can be separated by the constant multiple rule of limits.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\Bigg(e^{\displaystyle -x} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -h}-1}{-h}}$ $-$ $e^{\displaystyle x} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}\Bigg)}$

Let $h \,\to\, 0$, then $-h \,\to\, 0$. So, the limiting operator of the first limit can be evaluated as $-h$ approaches $0$.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\Bigg(e^{\displaystyle -x} \times \displaystyle \large \lim_{-h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -h}-1}{-h}}$ $-$ $e^{\displaystyle x} \times \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}\Bigg)}$

Now, the limit of (e^x-1)/x as x approaches 0 formula can be used to evaluate the limit of each function.

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\Big(e^{\displaystyle -x} \times 1$ $-$ $e^{\displaystyle x} \times 1\Big)$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$ $\times$ $\Big(e^{\displaystyle -x}-e^{\displaystyle x}\Big)$

$=\,\,\,$ $\dfrac{2 \times \Big(e^{\displaystyle -x}-e^{\displaystyle x}\Big)}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)^2}$

$=\,\,\,$ $\dfrac{2 \times \Big(e^{\displaystyle -x}-e^{\displaystyle x}\Big)}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times \Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$ $\times$ $\dfrac{\Big(e^{\displaystyle -x}-e^{\displaystyle x}\Big)}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$

$=\,\,\,$ $\dfrac{2}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$ $\times$ $\dfrac{-\Big(e^{\displaystyle x}-e^{\displaystyle -x}\Big)}{\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)}$

$=\,\,\,$ $\dfrac{2}{e^{\displaystyle x}+e^{\displaystyle -x}}$ $\times$ $\Bigg(-\dfrac{e^{\displaystyle x}-e^{\displaystyle -x}}{e^{\displaystyle x}+e^{\displaystyle -x}}\Bigg)$

The functions in natural exponential functions can be written in terms of hyperbolic secant and tangent functions.

$=\,\,\,$ $\operatorname{sech}{x}$ $\times$ $(-\tanh{x})$

$=\,\,\,$ $-\operatorname{sech}{x} \tanh{x}$

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

Thus, the derivative formula of the hyperbolic secant function is derived in differential calculus by the first principle of the differentiation.

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