Proof of Derivative of Hyperbolic Cosine function

Let $x$ represents a variable, the hyperbolic cosine function is written as $\cosh{x}$. The differentiation or the derivative of hyperbolic cosine function with respect to $x$ is written in below mathematical form.

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

In differential mathematics, the derivative formula of the hyperbolic cosine function can be derived by the first principle of the differentiation. Now, let us try to learn how to prove the differentiation rule of hyperbolic cosine function.

Derivative of Hyperbolic Cos function in Limit form

The derivative rule of hyperbolic cosine function can be proved in limit form by the fundamental definition of the derivative in differential calculus.

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

When $\Delta x$ is used to represent by $h$ simply, the whole expression in the right hand side of the equation is written in terms of $h$ instead of $\Delta x$.

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

The differentiation formula of $\cosh{(x)}$ function with respect to $x$ is actually derived in differential calculus mathematically from the first principle of differentiation.

Evaluate the Limit by the Direct Substitution

Firstly, try to evaluate the limit of the function by the direct substitution method as $h$ approaches zero for proving the derivative rule of the hyperbolic cosine function.

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

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

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

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

The direct substitution method is derived that the derivative of $\cosh{x}$ function is indeterminate. Actually, the differentiation of hyperbolic cosine function should not be indeterminate. Hence, the derivative rule of the hyperbolic cosine function must be proved in another method.

Simplify the Hyperbolic Sine function

The direct substitution method is not useful in this case for evaluating the limit of the function. Therefore, we should have to use an alternative approach for proving the differentiation of hyperbolic cosine function.

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

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

The expression in the numerator is in complex form. So, it is required to simplify this expression.

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

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

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

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

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

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

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

In the numerator, the first two terms can be expressed in their equivalent form by using the product rule of exponents with same base.

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

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

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

We can now take the common factors out from the terms in the expression of the numerator.

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

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

Now, use the constant multiple rule of limits to separate the constant from the function.

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

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

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

Use the sum rule of limits for evaluating the limit of sum of the functions by the sum of their limits.

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

The natural exponential functions $e^{\displaystyle x}$ and $e^{\displaystyle -x}$ are constants and they can be separated from the both terms.

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

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

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

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

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

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

Evaluate the Limit of the functions

The expression is simplified successfully and it is time to evaluate the limit of each function. If $h \,\to\, 0$, then $-h \,\to\, 0$. So, we can evaluate the limit of the second function as $-h$ approaches $0$.

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

The limit rule of (e^x-1)/x as x approaches 0 can be used to evaluate the limit of every 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 natural exponential functions is simply written as the hyperbolic sine function.

$= \,\,\,$ $\sinh{x}$

$\therefore \,\,\, \dfrac{d}{dx}{\, \cosh{x}} \,=\, \sinh{x}$

Thus, the derivative formula of hyperbolic cosine function can be derived by the first principle of the differentiation in differential calculus.