Derivative of Hyperbolic Sine using First Principle of Derivatives

We will be proving that the derivative of

\sinh(x)

is equal to

\cosh(x)

by using the first principle of derivatives.

Proof. Let

f(x) = \sinh(x) = \frac{e^x - e^{-x}}{2}

. Then by the first order principle, we have

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\sinh(x + h) - \sinh(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*}

As we know from this article, we have that

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

. The

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

is equal to -1, because 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}) = \cosh(x). \end{align*}

So we have

f'(x) = \cosh(x)

which completes the proof.