# Proof of Derivative of Hyperbolic Cotangent function

In mathematics, the hyperbolic cotangent function is expressed as $\coth{x}$ when $x$ represents a variable. The differentiation or the derivative of hyperbolic cot function with respect to $x$ is written in the following mathematical form.

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

The derivative of the hyperbolic cot function can be proved in differential calculus by the first principle of the differentiation. So, let’s learn how to derive the differentiation of hyperbolic cot function.

### Derivative of Hyperbolic Cot function in Limit form

The derivative of hyperbolic cotangent function can be derived in limit form in differential calculus by the fundamental definition of the derivative.

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

If $\Delta x$ is used to simply denote by $h$, then it can be expressed in $h$ instead of $\Delta x$.

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

By using the first principle of differentiation, the differentiation of $\coth{(x)}$ function with respect to $x$ can be proved in differential calculus.

### Evaluate the Limit by the Direct Substitution

Now, use the direct substitution method for evaluating the differentiation of hyperbolic cot function as $h$ approaches zero.

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

$= \,\,\,$ $\dfrac{\coth{x}-\coth{x}}{0}$

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

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

The derivative of the hyperbolic cot function does not equal to indeterminate form. So, direct substitution method is not useful in this case. So, we have to approach an alternative method to prove the derivative of $\coth{x}$ function.

### Simplify the Hyperbolic Cotangent function

Now, comeback to the expression of the derivative of hyperbolic cot function.

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

It can be simplified by expressing every hyperbolic cot function into its equivalent exponential expression form.

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

Now, let us concentrate on simplifying the expression.

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

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

The following expression is the expression in the numerator of the above mathematical expression.

$\Big(e^(x+h)}+e^-(x+h)}\Big) \Big(e^x}-e^-x}\Big$ $-$ $\Big(e^x}+e^-x}\Big) \Big(e^(x+h)}-e^-(x+h)}\Big$

Let’s simplify this mathematical expression and then we will replace it by its equivalent value.

$=\,\,\,$ $\Big($ $e^(x+h)} \times e^x$ $-$ $e^(x+h)} \times e^-x$ $+$ $e^-(x+h)} \times e^x$ $-$ $e^-(x+h)} \times e^-x$ $\Big)$ $-$ $\Big(e^x} \times e^(x+h)$ $-$ $e^x} \times e^-(x+h)$ $+$ $e^-x} \times e^(x+h)$ $-$ $e^-x} \times e^-(x+h)$ $\Big)$

$=\,\,\,$ $\Big($ $e^(x+h)+x$ $-$ $e^(x+h)-x$ $+$ $e^-(x+h)+x$ $-$ $e^-(x+h)-x$ $\Big)$ $-$ $\Big(e^x+(x+h)$ $-$ $e^x-(x+h)$ $+$ $e^-x+(x+h)$ $-$ $e^-x-(x+h)$ $\Big)$

$=\,\,\,$ $\Big($ $e^x+h+x$ $-$ $e^x+h-x$ $+$ $e^-x-h+x$ $-$ $e^-x-h-x$ $\Big)$ $-$ $\Big(e^x+x+h$ $-$ $e^x-x-h$ $+$ $e^-x+x+h$ $-$ $e^-x-x-h$ $\Big)$

$=\,\,\,$ $\Big($ $e^x+x+h$ $-$ $e^x-x+h$ $+$ $e^-x+x-h$ $-$ $e^-x-x-h$ $\Big)$ $-$ $\Big(e^x+x+h$ $-$ $e^x-x-h$ $+$ $e^-x+x+h$ $-$ $e^-x-x-h$ $\Big)$

$=\,\,\,$ $\require{cancel} \Big($ $e^2x+h$ $-$ $e^\cancel{x}-\cancel{x}+h$ $+$ $e^-\cancel{x}+\cancel{x}-h$ $-$ $e^-2x-h$ $\Big)$ $-$ $\Big(e^2x+h$ $-$ $e^\cancel{x}-\cancel{x}-h$ $+$ $e^-\cancel{x}+\cancel{x}+h$ $-$ $e^-2x-h$ $\Big)$

$=\,\,\,$ $\Big($ $e^2x+h$ $-$ $e^h$ $+$ $e^-h$ $-$ $e^-2x-h$ $\Big)$ $-$ $\Big(e^2x+h$ $-$ $e^-h$ $+$ $e^h$ $-$ $e^-2x-h$ $\Big)$

$=\,\,\,$ $e^2x+h$ $-$ $e^h$ $+$ $e^-h$ $-$ $e^-2x-h$ $-$ $e^2x+h$ $+$ $e^-h$ $-$ $e^h$ $+$ $e^-2x-h$

$=\,\,\,$ $e^2x+h$ $-$ $e^2x+h$ $-$ $e^h$ $-$ $e^h$ $+$ $e^-h$ $+$ $e^-h$ $-$ $e^-2x-h$ $+$ $e^-2x-h$

$=\,\,\,$ $\cancel{e^2x+h}$ $-$ $\cancel{e^2x+h}$ $-$ $2e^h$ $+$ $2e^-h$ $-$ $\cancel{e^-2x-h}$ $+$ $\cancel{e^-2x-h}$

$=\,\,\,$ $-2e^h}+2e^-h$

$=\,\,\,$ $-2\Big(e^h}-e^-h}\Big$

It is the simplified expression of the expression in the numerator. So, the expression in the numerator can be replaced by this expression.

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

### Evaluate the Limit of the function

If we try to evaluate the limit of the function by the direct substitution, the limit of the function becomes indeterminate. So, we have to split the function into two factors for proving the derivative of the hyperbolic cot function.

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

The limit of the product of functions can be evaluated by the product of their limits. So, use the product rule of the limits.

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

Now, evaluate the limit of the first function as $h$ approaches $0$ by the direct substitution method but the limit of second factor is not going to be evaluated at this time.

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

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

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

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

Now, let’s concentrate on evaluating the limit of the second function.

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

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

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

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

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

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

Use sum rule of limits to evaluate the limit of sum by the sum of their limits.

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

Let $h \,\to\, 0$, then $-h \,\to\, 0$. Therefore, $-h$ approaches $0$ when $h$ approaches $0$.

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

According to the limit rule of (e^x-1)/x as x approaches 0, the limit of each function is equal to one.

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

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

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

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

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

The function can be simplify denoted by the hyperbolic co-secant function.

$=\,\,\,$ $-\Big(\operatorname{csch}{x}\Big)^2$

$\therefore\,\,\,$ $\dfrac{d}{dx}{\, (\coth{x})}$ $\,=\,$ $-\operatorname{csch}^2{x}$

In this way, the derivative of hyperbolic cotangent function can derived in differential calculus from first principle of differentiation.

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