$x$ is a variable and the hyperbolic cosine function in terms of this variable is mathematically written as $\cosh{x}$. The differentiation or derivative of $\cosh{x}$ with respect to $x$ is expressed as $\dfrac{d}{dx}{\, \cosh{(x)}}$ in differential calculus. The differentiation of hyperbolic cosine with respect to $x$ can be written in limit form as per the fundamental definition of a derivative.
$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{ \normalsize \dfrac{f(x+h)-f(x)}{h}}$
If $f(x) \,=\, \cosh{x}$, then $f{(x+h)} \,=\, \cosh{(x+h)}$. It is time to substitute them to move ahead in deriving the derivative of hyperbolic cosine function with respect to $x$.
$\implies \dfrac{d}{dx}{\, \cosh{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\cosh{(x+h)}-\cosh{x}}{h}}$
The hyperbolic cosine functions $\cosh{x}$ and $\cosh{(x+h)}$ are written in terms of exponential terms according to the fundamental definition of hyperbolic cosine.
$\cosh{x} \,=\, \dfrac{e^x+e^{-x}}{2}$
$\cosh{(x+h)} \,=\, \dfrac{e^{(x+h)}+e^{-(x+h)}}{2}$
Now, substitute expansions of hyperbolic cosine functions $\cosh{x}$ and $\cosh{(x+h)}$ in the definition of the derivative of cosine function.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}+e^{-(x+h)}}{2}-\dfrac{e^x+e^{-x}}{2}}{h}}$
The exponential form functions are in complex form in the numerator. Now, it is time to simplify it further and it helps us to get the derivative of hyperbolic cosine function in the upcoming steps.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}+e^{-(x+h)}-e^x-e^{-x}}{2}}{h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}+e^{-(x+h)}-e^x-e^{-x}}{2}}{\dfrac{h}{1}}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{(x+h)}+e^{-(x+h)}-e^x-e^{-x}}{2} \times \dfrac{1}{h}}$
Multiply both fractions to simplify the function further.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{[e^{(x+h)}+e^{-(x+h)}-e^x-e^{-x}] \times 1}{2 \times h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{(x+h)}+e^{-(x+h)}-e^x-e^{-x}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x} \times e^{h}+e^{-x-h}-e^x-e^{-x}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}e^{h}+e^{-x}e^{-h}-e^{x}-e^{-x}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}e^{h}-e^{x}+e^{-x}e^{-h}-e^{-x}}{2h}}$
$e^x$ is a common factor in the first two terms and $e^{-x}$ is another common factor in the next two terms of the numerator. Now, take them common one after one.
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(e^{-h}-1)}}{2h}}$
$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{1}{2} \times \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(e^{-h}-1)}}{h}}$
$= \,\,\,$ $\dfrac{1}{2} \times \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(e^{-h}-1)}}{h}}$
$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(e^{-h}-1)}}{h}}$
$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \to 0}{\normalsize \Bigg[\dfrac{e^{x}{(e^{h}-1)}}{h}+\dfrac{e^{-x}{(e^{-h}-1)}}{h}}\Bigg]$
The limit of sum of functions can be written as sum of their limits as per sum law of limits.
$= \,\,\,$ $\dfrac{1}{2} \Bigg[\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}}{h}}$ $+$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-x}{(e^{-h}-1)}}{h}}\Bigg]$
$= \,\,\,$ $\dfrac{1}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}}{h}}$ $+$ $\dfrac{1}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-x}{(e^{-h}-1)}}{h}}$
The limit of each function is defined as $h$ tends to $0$. So, $e^x$ and $e^{-x}$ are common functions.
$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-h}-1}{h}}$
The expression in mathematical form is simplified successfully. Now, it is time to find the differentiation of hyperbolic cosine function.
$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-h}-1}{h} \times \dfrac{-1}{-1}}$
$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{(e^{-h}-1) \times -1}{h \times -1}}$
$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{-h \to 0}{\normalsize \dfrac{-(e^{-h}-1)}{-h}}$
$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $-$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{-h \to 0}{\normalsize \dfrac{e^{-h}-1}{-h}}$
According to exponential limit rule, the limit of every function is equal to one.
$= \,\,\,$ $\dfrac{e^{x}}{2} \times 1$ $-$ $\dfrac{e^{-x}}{2} \times 1$
$= \,\,\,$ $\dfrac{e^{x}}{2}-\dfrac{e^{-x}}{2}$
$= \,\,\,$ $\dfrac{e^{x}-e^{-x}}{2}$
$= \,\,\,$ $\sinh{x}$
$\therefore \,\,\, \dfrac{d}{dx}{\, \cosh{x}} \,=\, \sinh{x}$
Therefore, it is proved that the derivative of hyperbolic cosine with respect to $x$ is equal to hyperbolic sine in terms of $x$.
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