Maclaurin Series: Implicit Differentiation & Series Substitution — TJC 2025 H2 Math Prelim P2

(a) Establish the implicit relation and differentiate

Rewrite the logarithm of a square root, exponentiate to clear the log, then differentiate both sides.

\(y = \ln\sqrt{1+\mathrm{e}^{3x}} = \tfrac{1}{2}\ln(1+\mathrm{e}^{3x})\), so \(2y = \ln(1+\mathrm{e}^{3x})\), i.e. \(\mathrm{e}^{2y} = 1+\mathrm{e}^{3x}\).

Differentiating with respect to \(x\):

\[ 2\mathrm{e}^{2y}\frac{\mathrm{d}y}{\mathrm{d}x} = 3\mathrm{e}^{3x} \qquad \text{(Shown)} \tag{1} \]

(b) Maclaurin series for \(y\) up to \(x^2\)

Differentiate (1) again, then evaluate everything at \(x=0\) to get the derivative values.

Differentiating (1) with respect to \(x\):

\[ 2\mathrm{e}^{2y}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4\mathrm{e}^{2y}\!\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{\!2} = 9\mathrm{e}^{3x} \tag{2} \]

When \(x = 0\): \(y = \tfrac{1}{2}\ln 2\), so \(\mathrm{e}^{2y} = \mathrm{e}^{\ln 2} = 2.\)

Substituting into (1): \(2(2)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{4}\).

Substituting into (2): \(2(2)\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4(2)\!\left(\dfrac{3}{4}\right)^{\!2} = 9 \Rightarrow \dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \dfrac{9}{8}\).

Therefore:

\[ y = \frac{1}{2}\ln 2 + \frac{3}{4}x + \frac{9}{16}x^2 + \cdots \]

(c) Deduce the related series

Split off the \(\ln 2\), recognise the remaining log as the part (b) series with \(x\) replaced by \(-x\).

\[ \begin{aligned} \ln\!\left(\frac{\sqrt{1+\mathrm{e}^{-3x}}}{2}\right) &= \ln\!\left(\sqrt{1+\mathrm{e}^{3(-x)}}\right) - \ln 2\\ &= -\ln 2 + \frac{1}{2}\ln 2 + \frac{3}{4}(-x) + \frac{9}{16}(-x)^2 + \cdots\\ &= -\frac{1}{2}\ln 2 - \frac{3}{4}x + \frac{9}{16}x^2 + \cdots \end{aligned} \]