TMJC JC1 Promo 9758/2019/Q6

1. Express the equation in implicit form to avoid differentiating a natural logarithm.

Given $y = \ln(2 - \mathrm{e}^{2x})$
$\mathrm{e}^y = 2 - \mathrm{e}^{2x}$

2. Differentiate implicitly with respect to $x$ once.

$\mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^{2x}$

3. Differentiate implicitly a second time to show the required result.

Using the product rule on the left hand side:
$\mathrm{e}^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x}\right) \frac{\mathrm{d}y}{\mathrm{d}x} = -4\mathrm{e}^{2x}$
$\mathrm{e}^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \mathrm{e}^y \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -4\mathrm{e}^{2x}$
$\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -4\mathrm{e}^{2x}$ (Shown)

4. Evaluate $y$, $\frac{\mathrm{d}y}{\mathrm{d}x}$, and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ at $x=0$.

When $x = 0$:
$y = \ln(2 - \mathrm{e}^0) = \ln(1) = 0$

Substitute into the first derivative equation:
$\mathrm{e}^0 \frac{\mathrm{d}y}{\mathrm{d}x} = -2\mathrm{e}^0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -2$

Substitute into the second derivative equation:
$\mathrm{e}^0 \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (-2)^2 \right] = -4\mathrm{e}^0$
$1 \cdot \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 4 \right] = -4 \implies \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -8$

5. Differentiate the second derivative equation to find the third derivative at $x=0$.

Differentiate $\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -4\mathrm{e}^{2x}$ with respect to $x$:
$\mathrm{e}^y \left[ \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} \right] + \mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -8\mathrm{e}^{2x}$

Substitute $x=0, y=0, \frac{\mathrm{d}y}{\mathrm{d}x}=-2, \frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-8$:
$1 \cdot \left[ \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2(-2)(-8) \right] + 1 \cdot (-2) \left[ -8 + (-2)^2 \right] = -8$
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 32 - 2(-4) = -8$
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 40 = -8 \implies \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -48$

6. Construct the Maclaurin series using the evaluated values.

$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 0 + (-2)x + \frac{-8}{2}x^2 + \frac{-48}{6}x^3 + \dots$
$y = -2x - 4x^2 - 8x^3 + \dots$

1. Apply the binomial expansion to $(1-2x)^{\frac{1}{2}}$ up to the term in $x^2$.

2. Factorize to obtain the $(1+ax)^n$ form for $(2+3x)^{-1}$ and expand up to the term in $x^2$.

3. Multiply the two series together, ignoring terms with powers higher than $x^2$.

4. Determine the range of values of $x$ for which the expansion is valid.

The expansion for $(1-2x)^{\frac{1}{2}}$ is valid when:
$|-2x| < 1 \implies |x| < \frac{1}{2}$

The expansion for $\left(1 + \frac{3}{2}x\right)^{-1}$ is valid when:
$\left|\frac{3}{2}x\right| < 1 \implies |x| < \frac{2}{3}$

For the product expansion to be valid, both conditions must be satisfied. We take the intersection (the stricter condition):
$|x| < \frac{1}{2}$
$-\frac{1}{2} < x < \frac{1}{2}$