Maclaurin series: EJC 2024 Promo Q8

(a)

$$\begin{aligned} \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \sec^2\!\left[\ln(1+2x)\right]\times\frac{1}{1+2x}\times 2\\ &= \frac{2}{1+2x}\left\{1 + \tan^2\!\left[\ln(1+2x)\right]\right\}\\ &= \frac{2}{1+2x}\left(1 + y^2\right). \end{aligned}$$ Rearranging, $(1 + 2x)\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x} = 2\left(1 + y^2\right)$ (shown)

(b)

Differentiating the given result, $$\begin{aligned} (1 + 2x)\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} + \frac{\,\mathrm{d} y}{\,\mathrm{d} x}(2) &= 2(2y)\frac{\,\mathrm{d} y}{\,\mathrm{d} x}. \end{aligned}$$ When $x = 0$, $$y=\tan[\ln(1+0)]=\tan0=0,$$ $$\begin{aligned} (1+0)\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= 2(1+0) \;\Rightarrow\; \frac{\,\mathrm{d} y}{\,\mathrm{d} x} = 2,\\ (1+0)\frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} + 2(2) &= 2(0)(2) \;\Rightarrow\; \frac{\,\mathrm{d}^2 y}{\,\mathrm{d} x^2} = -4, \end{aligned}$$ so $$\begin{aligned} \tan\!\left[\ln(1+2x)\right] &= 0 + 2x + \frac{-4}{2!}x^2 + \cdots = 2x - 2x^2 + \cdots. \end{aligned}$$

(c)

$$\begin{aligned} \mathrm{e}^{\frac{1}{2}y} &= \mathrm{e}^{\frac{2x-2x^2+\cdots}{2}}\\ &= \mathrm{e}^{x - x^2 + \cdots}\\ &= 1 + \left(x - x^2 + \cdots\right) + \frac{\left(x - x^2 + \cdots\right)^2}{2!} + \cdots\\ &= 1 + x - \frac{1}{2}x^2 + \cdots. \end{aligned}$$

(d)

$$\begin{aligned} \lim_{x\to 0}\frac{\mathrm{e}^{\frac{1}{2}y} - 1}{3x} &= \lim_{x\to 0}\frac{\left(1 + x - \tfrac{1}{2}x^2 + \cdots\right) - 1}{3x}\\ &= \lim_{x\to 0}\frac{x - \tfrac{1}{2}x^2 + \cdots}{3x}\\ &= \lim_{x\to 0}\left(\frac{1}{3} - \frac{1}{6}x + \cdots\right)\\ &= \frac{1}{3}. \end{aligned}$$