Maclaurin series: DHS 2024 Promo Q5

(a)

$$\begin{aligned} \mathrm{e}^y&=2+\mathrm{e}^x \end{aligned}$$ Differentiate with respect to x $$\begin{aligned} \mathrm{e}^y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}&=\mathrm{e}^x \end{aligned}$$ Differentiate with respect to x $$\begin{aligned} \mathrm{e}^y\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right) \left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right) +\mathrm{e}^y\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}&=\mathrm{e}^x\\ \frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2 &=\frac{\mathrm{e}^x}{\mathrm{e}^y}\\ \frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2 &=\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\qquad\text{(shown)} \end{aligned}$$ When $x=0$, $y=\ln3$, $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}=\dfrac13$, $\dfrac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}=\dfrac29$ $$\begin{aligned} y&=\ln3+x\left(\frac13\right)+\frac{x^2}{2!}\left(\frac29\right)+\ldots\\ &=\ln3+\frac x3+\frac{x^2}{9}+\ldots \end{aligned}$$

(b)

$$\begin{aligned} y&=\ln(2+\mathrm{e}^x)\\ &=\ln\left(2+1+x+\frac{x^2}{2!}+\ldots\right)\\ &=\ln3\left[1+\frac13\left(x+\frac{x^2}{2!}+\ldots\right)\right]\\ &=\ln3+\ln\left[1+\frac13\left(x+\frac{x^2}{2!}+\ldots\right)\right] \end{aligned}$$ $$\begin{aligned} y=\ln(2+\mathrm{e}^x) &=\ln3+\ln[1+X] &&\left[\text{let }X=\frac13\left(x+\frac{x^2}{2!}+\ldots\right)\right]\\ &=\ln3+\left[X-\frac12X^2+\ldots\right] &&\text{( apply std series of $\ln(1+X)$ )}\\ &=\ln3+\left[\frac13\left(x+\frac{x^2}{2!}+\ldots\right) -\frac12\left(\frac13\left(x+\frac{x^2}{2!}+\ldots\right)\right)^2+\ldots\right]\\ &=\ln3+\left[\frac x3+\frac{x^2}{6}-\frac{x^2}{18}+\ldots\right]\\ &=\ln3+\frac x3+\frac{x^2}{9}+\ldots\quad\text{(verified)} \end{aligned}$$