Answer: (a) \(1-\tanh^2 u=\operatorname{sech}^2 u\).

(b) \(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1} t)=-\frac{1}{t\sqrt{1-t^2}}\).

(c) \(\frac{\mathrm{d}y}{\mathrm{d}x}=-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1} t\).

(d) \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=t\sqrt{1-t^2}-\frac{(1-t^2)^{3/2}}{t}-2t(1-t^2)\operatorname{sech}^{-1} t\).

(a) From the definitions,

\(\tanh u=\frac{\mathrm{e}^u-\mathrm{e}^{-u}}{\mathrm{e}^u+\mathrm{e}^{-u}}\), \(\operatorname{sech}u=\frac{2}{\mathrm{e}^u+\mathrm{e}^{-u}}\).

Then

\(1-\tanh^2u=1-\left(\frac{\mathrm{e}^u-\mathrm{e}^{-u}}{\mathrm{e}^u+\mathrm{e}^{-u}}\right)^2\)

\(=\frac{(\mathrm{e}^u+\mathrm{e}^{-u})^2-(\mathrm{e}^u-\mathrm{e}^{-u})^2}{(\mathrm{e}^u+\mathrm{e}^{-u})^2}\)

\(=\frac{4}{(\mathrm{e}^u+\mathrm{e}^{-u})^2}\)

\(=\left(\frac{2}{\mathrm{e}^u+\mathrm{e}^{-u}}\right)^2=\operatorname{sech}^2u\).

(b) Let \(u=\operatorname{sech}^{-1}t\). Then \(\operatorname{sech}u=t\).

Differentiating implicitly with respect to \(t\),

\(-\tanh u\,\operatorname{sech}u\,\frac{\mathrm{d}u}{\mathrm{d}t}=1\).

So

\(\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{1}{\tanh u\operatorname{sech}u}\).

Using part (a),

\(\tanh^2u=1-\operatorname{sech}^2u=1-t^2\).

Since \(0<t<1\), we take \(\tanh u=\sqrt{1-t^2}\), and \(\operatorname{sech}u=t\). Hence

\(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1}t)=\frac{\mathrm{d}u}{\mathrm{d}t}=-\frac{1}{t\sqrt{1-t^2}}\).

(c) Given \(x=\tanh^{-1}t\) and \(y=t\operatorname{sech}^{-1}t\),

\(\frac{\mathrm{d}y}{\mathrm{d}t}=\operatorname{sech}^{-1}t+t\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1}t)=\operatorname{sech}^{-1}t-\frac{1}{\sqrt{1-t^2}}\).

Also

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{1-t^2}\).

Therefore

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\cdot\frac{\mathrm{d}t}{\mathrm{d}x}=\left(\operatorname{sech}^{-1}t-\frac{1}{\sqrt{1-t^2}}\right)(1-t^2)\)

\(=-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1}t\).

(d) Differentiate the result from part (c) with respect to \(t\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(-\sqrt{1-t^2}+(1-t^2)\operatorname{sech}^{-1}t\right)\)

\(=\frac{t}{\sqrt{1-t^2}}-\frac{1-t^2}{t\sqrt{1-t^2}}-2t\operatorname{sech}^{-1}t\).

Now

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot\frac{\mathrm{d}t}{\mathrm{d}x}\), and since \(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{1-t^2}\), we have \(\frac{\mathrm{d}t}{\mathrm{d}x}=1-t^2\).

So

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=(1-t^2)\left(\frac{t}{\sqrt{1-t^2}}-\frac{1-t^2}{t\sqrt{1-t^2}}-2t\operatorname{sech}^{-1}t\right)\)

\(=t\sqrt{1-t^2}-\frac{(1-t^2)^{3/2}}{t}-2t(1-t^2)\operatorname{sech}^{-1}t\).