Answer: (a) \(\cosh^2 x-\sinh^2 x=1\).

(b) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\big(\tan^{-1}(\sinh x)\big)=\operatorname{sech}x\).

(c) The graph of \(y=\operatorname{sech}x\) is an even curve, symmetric about the \(y\)-axis, with maximum value \(1\) at \(x=0\), and horizontal asymptote \(y=0\).

(d) \(\displaystyle \sum_{r=1}^{n}\operatorname{sech}r\lt \tan^{-1}(\sinh n)\).

(e) An upper bound for \(\displaystyle \sum_{r=1}^{\infty}\operatorname{sech}r\) is \(\displaystyle \frac{\pi}{2}\).

(a) Using the definitions

\(\cosh x=\dfrac{e^x+e^{-x}}{2}, \qquad \sinh x=\dfrac{e^x-e^{-x}}{2}.\)

Then

\(\cosh^2x=\left(\dfrac{e^x+e^{-x}}{2}\right)^2=\dfrac{e^{2x}+2+e^{-2x}}{4}\)

and

\(\sinh^2x=\left(\dfrac{e^x-e^{-x}}{2}\right)^2=\dfrac{e^{2x}-2+e^{-2x}}{4}.\)

So

\(\cosh^2x-\sinh^2x=\dfrac{e^{2x}+2+e^{-2x}}{4}-\dfrac{e^{2x}-2+e^{-2x}}{4}=\dfrac{4}{4}=1.\)

Hence \(\cosh^2 x-\sinh^2 x=1\).

(b) Let \(u=\sinh x\). Then

\(\dfrac{\mathrm{d}}{\mathrm{d}x}\big(\tan^{-1}(\sinh x)\big)=\dfrac{1}{1+u^2}\cdot \dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\cosh x}{1+\sinh^2x}.\)

From part (a), \(\cosh^2x-\sinh^2x=1\), so \(1+\sinh^2x=\cosh^2x\). Therefore

\(\dfrac{\mathrm{d}}{\mathrm{d}x}\big(\tan^{-1}(\sinh x)\big)=\dfrac{\cosh x}{\cosh^2x}=\dfrac{1}{\cosh x}=\operatorname{sech}x.\)

(c) Since \(\operatorname{sech}x=\dfrac{1}{\cosh x}\) and \(\cosh x\) is even, \(\operatorname{sech}x\) is also even, so the graph is symmetric about the \(y\)-axis.

Also, \(\cosh 0=1\), so \(\operatorname{sech}0=1\). This is the maximum point.

As \(|x|\to\infty\), \(\cosh x\to\infty\), so \(\operatorname{sech}x\to 0\). Hence the horizontal asymptote is \(y=0\).

So the sketch is a positive curve, peaking at \((0,1)\), symmetric about the \(y\)-axis, and approaching \(0\) as \(x\to \pm\infty\).

(d) On \([0,\infty)\), the graph of \(y=\operatorname{sech}x\) decreases. Using rectangles of width \(1\) and heights \(\operatorname{sech}1,\operatorname{sech}2,\dots,\operatorname{sech}n\), each rectangle lies below the curve over the interval to its left.

Hence

\(\displaystyle \sum_{r=1}^{n}\operatorname{sech}r\lt \int_0^n \operatorname{sech}x\,\mathrm{d}x.\)

From part (b), an antiderivative of \(\operatorname{sech}x\) is \(\tan^{-1}(\sinh x)\). So

\(\displaystyle \int_0^n \operatorname{sech}x\,\mathrm{d}x=\left[\tan^{-1}(\sinh x)\right]_0^n=\tan^{-1}(\sinh n)-\tan^{-1}(0)=\tan^{-1}(\sinh n).\)

Therefore

\(\displaystyle \sum_{r=1}^{n}\operatorname{sech}r\lt \tan^{-1}(\sinh n).\)

(e) As \(n\to\infty\), \(\sinh n\to\infty\), so \(\tan^{-1}(\sinh n)\to \dfrac{\pi}{2}.\)

Hence \(\displaystyle \sum_{r=1}^{\infty}\operatorname{sech}r\) has upper bound \(\displaystyle \frac{\pi}{2}\).