Find the derivative of Hyperbolic Tangent

The Mathematician — Mon, 17 Oct 2022 13:00:04 +0000

The derivative of

\tanh(x)

\text{sech}^2(x)

.

Solution. Let

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

We will use the quotient rule, where we let

\begin{align*} \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{g(x)}{h(x)}, \end{align*}

such that we can determine

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

We know that

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

and that

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

. So we get

\begin{align*} g'(x) = e^x + e^{-x} \quad \text{and} \quad h'(x) = e^x - e^{-x}. \end{align*}

So wrapping everything together, we get

\begin{align*} f'(x) &= \frac{d}{dx} \tanh(x) \\ &= \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \\ &= \frac{(e^x + e^{-x})(e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2} \\ &= \frac{e^{2x} + e^{-2x} + 2 - (e^{2x} - e^{-2x} - 2 )}{(e^x + e^{-x})^2} \\ &= \frac{4}{(e^x + e^{-x})^2} \\ &= \frac{2^2}{(e^x + e^{-x})^2} \\ &= \text{sech}^2(x). \end{align*}

So, the derivative of

\tanh(x)

\text{sech}^2(x)

Derivative of Hyperbolic Tangent using First Principle of Derivatives

The Mathematician — Sat, 15 Oct 2022 13:00:08 +0000

We have seen earlier the derivatives of

\sinh(x)

and

\cosh(x)

using the first principle of derivatives. We will prove that, too, for the

\tanh(x)

using the first principle of derivatives.

Proof. Let

f(x) = \tanh(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{\tanh(x + h) - \tanh(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*}

Now what is left to prove is

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

. Notice that

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*}

We have seen in this article, that we have

\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*} \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{sech}^2(x). \end{align*}