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Derivative of Hyperbolic Secant using First Principle of Derivatives

Derivative of Hyperbolic Secant using First Principle of Derivatives

The derivative of sech(x)\text{sech}(x) is −sech(x)tanh⁡(x)-\text{sech}(x)\tanh(x). We will prove that with the first principle of derivatives.

Proof. Let f(x)=sech(x)=2ex+e−xf(x) = \text{sech}(x) = \frac{2}{e^x + e^{-x}}. Then using the first principle of derivatives, we get:
f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0sech(x+h)−sech(x)h=lim⁡h→02ex+h+e−x−h−2(ex+e−x)(ex+h+e−x−h)h=lim⁡h→02(ex+e−x)−2(ex+h+e−x−h)ex+h+e−x−hh=2lim⁡h→0ex+e−x−ex+h−e−x−h(ex+h+e−x−h)(ex+e−x)h=2lim⁡h→0ex+e−x−ex+h−e−x−hh(ex+h+e−x−h)(ex+e−x)=2lim⁡h→0ex(1−eh)+e−x(1−e−h)h(ex+h+e−x−h)(ex+e−x)=2lim⁡h→0ex(1−eh)h(ex+h+e−x−h)(ex+e−x)+2lim⁡h→0e−x(1−e−h)h(ex+h+e−x−h)(ex+e−x)=2exlim⁡h→01−ehh(ex+h+e−x−h)(ex+e−x)+2e−xlim⁡h→01−e−hh(ex+h+e−x−h)(ex+e−x)\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:
2exlim⁡h→01−ehh(ex+h+e−x−h)(ex+e−x)=2exlim⁡h→01−ehh⋅lim⁡h→01(ex+h+e−x−h)(ex+e−x)=2exlim⁡h→01−ehh⋅1(ex+e−x)2=2ex(ex+e−x)2lim⁡h→01−ehh.\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→01−ehh=−lim⁡h→0eh−1h=−1\lim_{h \rightarrow 0} \frac{1 - e^{h}}{h} = - \lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} = -1. So we get
f′(x)=−2ex(ex+e−x)2+2e−xlim⁡h→01−e−hh(ex+h+e−x−h)(ex+e−x).\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:
2e−xlim⁡h→01−e−hh(ex+h+e−x−h)(ex+e−x)=2e−x(ex+e−x)2lim⁡h→01−e−hh\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→01−e−hh=−lim⁡h→0e−h−1h=1\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:
f′(x)=−2ex(ex+e−x)2+2e−x(ex+e−x)2=−2(ex−e−x)(ex+e−x)2=−sech(x)ex−e−xex+e−x=−sech(x)tanh⁡(x).\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 sech(x)\text{sech}(x) is −sech(x)tanh⁡(x)-\text{sech}(x)\tanh(x).
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