Answer: (a) \(\cosh y=\dfrac{1}{x+\frac12}\), and hence

\(\sinh y\,\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{1}{\left(x+\frac12\right)^2}.\)

(b) The first three terms of the Maclaurin series are

\(\operatorname{sech}^{-1}\left(x+\frac12\right)=\ln(2+\sqrt3)-\dfrac{4}{\sqrt3}x+\dfrac{8}{3\sqrt3}x^2.\)

So \(a=2+\sqrt3\), \(b=-\dfrac{4}{\sqrt3}\), \(c=\dfrac{8}{3\sqrt3}\).

Let \(y=\operatorname{sech}^{-1}\left(x+\frac12\right)\).

(a) Since \(\operatorname{sech} y=x+\frac12\),

\(\dfrac{1}{\cosh y}=x+\frac12\),

so

\(\cosh y=\dfrac{1}{x+\frac12}=\left(x+\frac12\right)^{-1}.\)

Differentiating both sides with respect to \(x\),

\(\dfrac{\mathrm{d}}{\mathrm{d}x}(\cosh y)=\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x+\frac12\right)^{-1}.\)

Hence

\(\sinh y\,\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\left(x+\frac12\right)^{-2}=-\dfrac{1}{\left(x+\frac12\right)^2}.\)

(b) To find the Maclaurin expansion up to \(x^2\), we need \(y(0)\), \(y'(0)\) and \(y''(0)\).

From part (a),

\(\sinh y\,y'=-\left(x+\frac12\right)^{-2}.\)

Differentiating again,

\(\cosh y\,(y')^2+\sinh y\,y''=2\left(x+\frac12\right)^{-3}.\)

Now evaluate at \(x=0\).

First,

\(y(0)=\operatorname{sech}^{-1}\left(\frac12\right).\)

If \(\operatorname{sech} y=\frac12\), then \(\cosh y=2\), so

\(y(0)=\cosh^{-1}(2)=\ln(2+\sqrt3).\)

Also, when \(x=0\), \(\cosh y=2\), so

\(\sinh y=\sqrt{\cosh^2 y-1}=\sqrt{4-1}=\sqrt3.\)

Using \(\sinh y\,y'=-\left(x+\frac12\right)^{-2}\) at \(x=0\),

\(\sqrt3\,y'(0)=-\left(\frac12\right)^{-2}=-4,\)

so

\(y'(0)=-\dfrac{4}{\sqrt3}.\)

Now use

\(\cosh y\,(y')^2+\sinh y\,y''=2\left(x+\frac12\right)^{-3}.\)

At \(x=0\),

\(2\left(\dfrac{16}{3}\right)+\sqrt3\,y''(0)=2\left(\frac12\right)^{-3}=16.\)

Thus

\(\dfrac{32}{3}+\sqrt3\,y''(0)=16,\)

so

\(\sqrt3\,y''(0)=\dfrac{16}{3},\)

and therefore

\(y''(0)=\dfrac{16}{3\sqrt3}.\)

Hence the Maclaurin series is

\(y=y(0)+y'(0)x+\dfrac{y''(0)}{2}x^2,\)

so

\(\operatorname{sech}^{-1}\left(x+\frac12\right)=\ln(2+\sqrt3)-\dfrac{4}{\sqrt3}x+\dfrac{8}{3\sqrt3}x^2.\)

Therefore

\(a=2+\sqrt3,\quad b=-\dfrac{4}{\sqrt3},\quad c=\dfrac{8}{3\sqrt3}.\)