Math Doubts

Proof of Derivative of Hyperbolic Cosecant function

In mathematics, the hyperbolic cosecant function is written as $\operatorname{csch}{x}$ if $x$ represents a variable. The differentiation or the derivative of hyperbolic cosecant function with respect to $x$ is written in the following mathematical form.

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

In differential calculus, the derivative of the hyperbolic cosecant function is derived by the first principle of the differentiation. So, let us learn how to prove the differentiation formula for the hyperbolic cosecant function.

Derivative of Hyperbolic Cosecant in Limit form

The fundamental definition of the derivative is used to derive the differentiation of hyperbolic cosecant function. According to first principle of the differentiation, the derivative of hyperbolic cosecant function $\operatorname{csch}{(x)}$ can be expressed in limit form.

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

Now, let us assume that $\Delta x$ is denoted by $h$ simply. Therefore, the above equation can be written in terms of $h$ instead of $\Delta x$.

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

Evaluate the Limit by the Direct Substitution

Now, let’s observe that what happens when we use the direct substitution method for evaluating the differentiation of hyperbolic cosecant function as $h$ closer to zero.

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

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

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

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

It is derived that the derivative of hyperbolic cosecant function is indeterminate. It clears that it is not possible to use the direct substitution method for proving the differentiation formula for hyperbolic cosecant function. Hence, we should have to think about an alternative method.

Simplify the Hyperbolic Cosecant function

Now, focus on the definition of derivative of the cosecant function in limit form.

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

For simplifying the mathematical expression in the right hand side of the equation, express each hyperbolic cosecant function in terms of natural exponential functions.

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

We can now focus on simplifying the mathematical expression in the right hand side of the equation for proving the differentiation law of hyperbolic cosecant function.

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

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

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

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

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

The expression in the numerator can be simplified by using the product rule of exponents to third and fourth terms of the second factor.

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

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

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

The function in the above expression can be split as the product of two functions.

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

The limit of the product can be evaluated 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

We have two multiplying functions and let us evaluate the limit of each function. Now, find the limit of the first factor by the direct substitution method.

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

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

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

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

Now, let us evaluate the limit of the second multiplying function but we cannot use direct substitution method for the second factor because it becomes indeterminate if we try it. So, let’s use another mathematical approach.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The limit of sum of the functions can be evaluated by the sum rule of limits.

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

The constant multiple rule of limits can be used for separating the constants from the limiting operations.

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

If $h \,\to\, 0$, then $-h \,\to\, 0$. Therefore, the input of limiting operation of the second function can be evaluated as $-h$ approaches $0$.

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

The limit of each function in exponential form can be evaluated by the limit of (e^x-1)/x as x approaches 0 rule.

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

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

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

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

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

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

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

The above functions in natural exponential functions can be simply expressed by the hyperbolic co-secant and co-tangent functions.

$=\,\,\,$ $(-\operatorname{csch}{x})$ $\times$ $\coth{x}$

$=\,\,\,$ $-\operatorname{csch}{x} \coth{x}$

$\therefore\,\,\,$ $\dfrac{d}{dx}{\, (\operatorname{csch}{x})}$ $\,=\,$ $-\operatorname{csch}{x}\coth{x}$

In differential calculus, the derivative rule of the hyperbolic cosecant function is derived in the above way by the first principle of the differentiation.

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