The hyperbolic sine function is written as $\sinh{x}$ when $x$ is used to represent a variable. The differentiation or the derivative of hyperbolic sin function with respect to $x$ is written in the following mathematical form.
$\dfrac{d}{dx}{\, \Big(\sinh{(x)}\Big)}$
Mathematically, the differentiation formula of the hyperbolic sine function can be proved in differential calculus by the first principle of the differentiation. Now, let us learn how to derive the differentiation rule of hyperbolic sine function.
In differential calculus, the differentiation rule of hyperbolic sine function is derived in limit form by the fundamental definition of the derivative.
$\dfrac{d}{dx}{\, (\sinh{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\sinh{(x+\Delta x)}-\sinh{x}}{\Delta x}}$
If we take $\Delta x$ is denoted by $h$, then the whole mathematical expression can be written in terms of $h$ instead of $\Delta x$.
$\implies$ $\dfrac{d}{dx}{\, (\sinh{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sinh{(x+h)}-\sinh{x}}{h}}$
The derivative formula of $\sinh{(x)}$ function with respect to $x$ can be derived mathematically in differential calculus from the first principle of differentiation.
Let’s first try to evaluate the limit of the function by the direct substitution method as $h$ approaches zero for deriving the derivative formula of the hyperbolic sine function.
$= \,\,\,$ $\dfrac{\sinh{(x+0)}-\sinh{x}}{0}$
$= \,\,\,$ $\dfrac{\sinh{x}-\sinh{x}}{0}$
$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\sinh{x}}-\cancel{\sinh{x}}}{0}$
$=\,\,\,$ $\dfrac{0}{0}$
The method of direct substitution is derived that the derivative of the $\sinh{x}$ function is indeterminate. In fact, the differentiation of hyperbolic sine function never be indeterminate. Therefore, the derivative law of the hyperbolic sine function should be derived in differentiation in another mathematical approach.
It is not recommendable to evaluate the limit of the function by the direct substitution method. Hence, we have to think about the alternative method to evaluate it for deriving the proof of the derivative of hyperbolic sine function.
$\implies$ $\dfrac{d}{dx}{\, (\sinh{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sinh{(x+h)}-\sinh{x}}{h}}$
$\implies$ $\dfrac{d}{dx}{\, (\sinh{x})}$ $\,=\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}}{2}-\dfrac{e^{\displaystyle x}-e^{\displaystyle -x}}{2}}{h}}$
The functions in the numerator is in complex form and it is required to simplify the expression in the right hand side of the equation.
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-\Big(e^{\displaystyle x}-e^{\displaystyle -x}\Big)}{2}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-e^{\displaystyle x}+e^{\displaystyle -x}}{2}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-e^{\displaystyle x}+e^{\displaystyle -x}}{2}}{\dfrac{h}{1}}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-e^{\displaystyle x}+e^{\displaystyle -x}}{2} \times \dfrac{1}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-e^{\displaystyle x}+e^{\displaystyle -x}\Big) \times 1}{2 \times h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle (x+h)}-e^{\displaystyle -(x+h)}-e^{\displaystyle x}+e^{\displaystyle -x}}{2h}}$
The first two terms in the numerator can be written in their equivalent form by the product rule of exponents with same base.
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x} \times e^{\displaystyle h}-e^{\displaystyle -x-h}-e^{\displaystyle x}+e^{\displaystyle -x}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}e^{\displaystyle h}-e^{\displaystyle -x}e^{\displaystyle -h}-e^{\displaystyle x}+e^{\displaystyle -x}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}e^{\displaystyle h}-e^{\displaystyle x}-e^{\displaystyle -x}e^{\displaystyle -h}+e^{\displaystyle -x}}{2h}}$
Now, take the common factors out from the terms in the expression of the numerator.
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}+e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{2} \times \dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}+e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{h}\Bigg)}$
Separate the constant from the function by the constant multiple rule of limits.
$= \,\,\,$ $\dfrac{1}{2} \times \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}+e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{h}}$
$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}+e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{h}}$
$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}}{h}+\dfrac{e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{h}\Bigg)}$
According to sum rule of limits, the limit of sum of functions can be written as sum of their limits.
$= \,\,\,$ $\dfrac{1}{2} \Bigg(\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}{(e^{\displaystyle h}-1)}}{h}}$ $+$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -x}{(-e^{\displaystyle -h}+1)}}{h}\Bigg)}$
$e^{\displaystyle x}$ and $e^{\displaystyle -x}$ are constants. So, they can be separated from the both terms.
$= \,\,\,$ $\dfrac{1}{2}\Bigg(e^{\displaystyle x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}}$ $+$ $e^{\displaystyle -x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-e^{\displaystyle -h}+1}{h}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2}\Bigg(e^{\displaystyle x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}}$ $+$ $e^{\displaystyle -x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-\Big(e^{\displaystyle -h}-1\Big)}{h}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2}\Bigg(e^{\displaystyle x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}}$ $+$ $e^{\displaystyle -x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -h}-1}{-h}\Bigg)}$
The mathematical expression is successfully simplified and it is time to find the limit of each function. Let $h \,\to\, 0$, then $-h \,\to\, 0$. Therefore, we can find the limit of the second function as $-h$ approaches $0$.
$= \,\,\,$ $\dfrac{1}{2}\Bigg(e^{\displaystyle x}\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle h}-1}{h}}$ $+$ $e^{\displaystyle -x}\large \displaystyle \lim_{-h \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle -h}-1}{-h}\Bigg)}$
Use, the limit rule of (e^x-1)/x as x approaches 0 function to evaluate limit of the each function.
$= \,\,\,$ $\dfrac{1}{2}\Big(e^{\displaystyle x} \times 1$ $+$ $e^{\displaystyle -x} \times 1\Big)$
$= \,\,\,$ $\dfrac{1}{2}\Big(e^{\displaystyle x}+e^{\displaystyle -x}\Big)$
$= \,\,\,$ $\dfrac{e^{\displaystyle x}+e^{\displaystyle -x}}{2}$
The expression in terms of natural exponential functions is simply expressed as the hyperbolic cosine function.
$= \,\,\,$ $\cosh{x}$
$\therefore \,\,\, \dfrac{d}{dx}{\, \sinh{x}} \,=\, \cosh{x}$
In this way, the derivative of hyperbolic sine function is derived from the first principle of the differentiation.
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