What is the Derivative of Hyperbolic Cotangent?

The Mathematician — Mon, 24 Oct 2022 13:00:39 +0000

The derivative of

\coth(x)

-\text{csch}^2(x)

.

Solution. We know that

f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}

. So there are two different ways to prove this. The first way is to do it via

f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{g(x)}{h(x)}

. We will apply the quotient rule here:

\begin{align*} f'(x) = \frac{d}{dx} \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{g(x)}{h(x)} = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}. \end{align*}

Since

g'(x) = \frac{d}{dx} \cosh(x) = \sinh(x)

and

h'(x) = \frac{d}{dx} \sinh(x) = \cosh(x)

, we have that

\begin{align*} f'(x) &= \frac{d}{dx} \coth(x) \\ &= \frac{\cosh(x)}{\sinh(x)} \\ &= \frac{\sinh(x)\sinh(x) - \cosh(x)\cosh(x)}{\sinh^2(x)} \\ &= \frac{\sinh^2(x) - \cosh^2(x)}{\sinh^2(x)} \\ &= \frac{-1}{\sinh^2(x)} \\ &= -\text{csch}^2(x). \end{align*}

So the derivative of

\coth(x)

-\text{csch}^2(x)

. The second way is to do it by

f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}

, which we leave to the reader.

Derivative of Hyperbolic Cotangent using First Principle of Derivatives

The Mathematician — Thu, 20 Oct 2022 13:00:35 +0000

We will see that the derivative of

\coth(x)

-\text{csch}^2(x)

by using the first principle of derivatives.

Proof. Let

f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}

. Then

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

We still need to figure out what

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

is. We can apply an easy trick:

- e^{h} + e^{-h} = e^{h} - 1 + 1 - e^{-h}

. So

\begin{align*} \lim_{h \rightarrow 0} \frac{e^{h} - e^{-h}}{h} &= \lim_{h \rightarrow 0} \frac{- e^{h} - 1 + 1 + e^{-h}}{h} \\ &= - \lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} + \lim_{h \rightarrow 0} \frac{e^{-h} - 1}{h} \end{align*}

Because

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

which we have seen in article, and that

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

is equal to -1, as we could substitute

-h

in the Maclaurin series, we do have:

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

Continuing where we have been, we get

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

Therefore, we get

\begin{align*} f'(x) = -\frac{2^2}{(e^{x} - e^{-x})^2} = -\text{csch}^2(x) \end{align*}

which completes the proof.