Derivative of Hyperbolic Cosine using First Principle of Derivatives

In this article, we will find the derivative of

\cosh(x)

using the first principle of derivatives.

Proof. Let

f(x) = \cosh(x)

. We know that

\cosh(x)

is equal to:

\begin{align*} \cosh(x) = \frac{e^x + e^{-x}}{2}. \end{align*}

We want to determine the next limit:

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\cosh(x + h) - \cosh(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\frac{e^{x + h} + e^{-x-h}}{2} - \frac{e^x + e^{-x}}{2}}{h} \\ &= \lim_{h \rightarrow 0} \frac{e^{x + h} + e^{-x-h} - e^x - e^{-x}}{2h} \\ &= \lim_{h \rightarrow 0} \frac{e^{x}(e^{h} - 1) + e^{-x}(e^{-h} - 1)}{2h} \\ &= \frac{1}{2}e^x\lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} + \frac{1}{2}e^{-x}\lim_{h \rightarrow 0} \frac{e^{-h} - 1}{h} \end{align*}

Now we do know that from article that

\lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} = 1

. The

\lim_{h \rightarrow 0} \frac{e^{-h} - 1}{h}

is a little twist from the previous one, where we could substitute

-h

in the Maclaurin series. See the mentioned article. Therefore, we get

\begin{align*} \frac{1}{2}e^x\lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} + \frac{1}{2}e^{-x}\lim_{h \rightarrow 0} \frac{e^{-h} - 1}{h} = \frac{1}{2}(e^x - e^{-x}) = \sinh(x). \end{align*}

Therefore, as conclusion, we have

f'(x) = \sinh(x)