Derivative of Hyperbolic Secant using First Principle of Derivatives

The derivative of

\text{sech}(x)

-\text{sech}(x)\tanh(x)

. We will prove that with the first principle of derivatives.

Proof. Let

f(x) = \text{sech}(x) = \frac{2}{e^x + e^{-x}}

. Then using the first principle of derivatives, we get:

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

Let’s check the first part:

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

We have seen here that

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

. So we get

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

For the second part, we have:

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

Again, we saw here that

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

if we apply the Maclaurin series. So we get together:

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

So we get indeed that the derivative of

\text{sech}(x)

-\text{sech}(x)\tanh(x)