Maclaurin series: ACJC 2025 Promo Q2

Squaring $y = \sqrt{1 + \tan^{-1}2x}$ gives $y^2 = 1 + \tan^{-1}2x$. Differentiating: $$\begin{aligned} 2y\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \frac{2}{1+4x^2}\\ y\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= (1+4x^2)^{-1}. \end{aligned}$$ Differentiating again (product rule on the left): $$\begin{aligned} y\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} + \left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^{\!2} &= -(1+4x^2)^{-2}(8x)\\ y\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} + \left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^{\!2} + \frac{8x}{(1+4x^2)^2} &= 0. \quad\text{((shown))} \end{aligned}$$ Differentiating once more: $$\begin{aligned} y\frac{\,\mathrm{d}^3 y}{\,\mathrm{d} x^3} + 3\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} + 8(1+4x^2)^{-2} - 128x^2(1+4x^2)^{-3} &= 0. \end{aligned}$$ At $x = 0$: $y = 1$. From $y\,y' = (1+4x^2)^{-1}$: $y' = 1$. From the second relation: $y'' = -1$. From the third relation: $$\begin{aligned} y''' + 3(1)(-1) + 8 &= 0\\ y''' &= -5. \end{aligned}$$ Hence the Maclaurin series is $$\begin{aligned} y &\approx 1 + x - \frac{1}{2!}x^2 - \frac{5}{3!}x^3\\ &= 1 + x - \frac{1}{2}x^2 - \frac{5}{6}x^3. \end{aligned}$$