What is the Derivative of Hyperbolic Sine?

The Mathematician — Fri, 14 Oct 2022 13:00:55 +0000

The derivative of

\sinh(x)

\cosh(x)

.

Solution. Let

f(x) = \sinh(x)

. We know that

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

and that

\frac{d}{dx} e^x = e^x

and

\frac{d}{dx} e^{-x} = -e^{-x}

. So we get

\begin{align*} f'(x) &= \frac{d}{dx} \sinh(x) \\ &= \frac{d}{dx} \frac{e^x - e^{-x}}{2} \\ &= \frac{d}{dx} \frac{e^x}{2} - \frac{d}{dx} \frac{e^{-x}}{2} \\ &= \frac{e^x}{2} - \frac{-e^{-x}}{2} \\ &= \frac{e^x}{2} + \frac{e^{-x}}{2} \\ &= \frac{e^x + e^{-x}}{2} \\ &= \cosh(x). \end{align*}

In conclusion, we have that the derivative of

\sinh(x)

\cosh(x)

Derivative of Hyperbolic Sine using First Principle of Derivatives

The Mathematician — Sat, 08 Oct 2022 13:00:52 +0000

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.