2020/Promo/SAJC/Q3

1. Apply the product rule for differentiation

Let $y = e^{2x}\cos^{-1}(3x)$.
Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, we setup the differentiation:
$\frac{dy}{dx} = \left[\frac{d}{dx}\left(e^{2x}\right)\right]\cos^{-1}(3x) + e^{2x}\left[\frac{d}{dx}\left(\cos^{-1}(3x)\right)\right]$

2. Differentiate the individual terms using the chain rule

Differentiate the exponential term:
$\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}$

Differentiate the inverse trigonometric term using the standard derivative for $\cos^{-1}(x)$ alongside the chain rule:
$\frac{d}{dx}\left(\cos^{-1}(3x)\right) = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot \frac{d}{dx}(3x) = -\frac{3}{\sqrt{1 - 9x^2}}$

3. Combine and simplify the expression

Substitute the derivatives back into the product rule expression and factorise:
$\frac{dy}{dx} = 2e^{2x}\cos^{-1}(3x) + e^{2x}\left(-\frac{3}{\sqrt{1 - 9x^2}}\right)$
$= e^{2x} \left( 2\cos^{-1}(3x) - \frac{3}{\sqrt{1 - 9x^2}} \right)$

1. Differentiate the given equation implicitly with respect to $x$ to find $\frac{dy}{dx}$.

$\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2}\frac{dy}{dx} = 0$\n$\frac{dy}{dx} = -\frac{x^{-1/2}}{y^{-1/2}} = -x^{-1/2}y^{1/2}$

2. Differentiate $\frac{dy}{dx}$ using the product rule to find $\frac{d^2y}{dx^2}$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-x^{-1/2}y^{1/2}\right)$\n$\frac{d^2y}{dx^2} = -\left( -\frac{1}{2}x^{-3/2}y^{1/2} + x^{-1/2}\left(\frac{1}{2}y^{-1/2}\frac{dy}{dx}\right) \right)$\n$\frac{d^2y}{dx^2} = \frac{1}{2}x^{-3/2}y^{1/2} - \frac{1}{2}x^{-1/2}y^{-1/2}\frac{dy}{dx}$

3. Substitute the expression for $\frac{dy}{dx}$ into $\frac{d^2y}{dx^2}$ to express it in terms of $x$ and $y$ only.

Substitute $\frac{dy}{dx} = -x^{-1/2}y^{1/2}$ into the second derivative:\n$\frac{d^2y}{dx^2} = \frac{1}{2}x^{-3/2}y^{1/2} - \frac{1}{2}x^{-1/2}y^{-1/2}\left(-x^{-1/2}y^{1/2}\right)$\n$\frac{d^2y}{dx^2} = \frac{1}{2}x^{-3/2}y^{1/2} + \frac{1}{2}x^{-1}$

4. Substitute the expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the left-hand side (LHS) of the target equation to complete the proof.

$\text{LHS} = 2x\frac{d^2y}{dx^2} + \frac{dy}{dx} - 1$\n$= 2x\left(\frac{1}{2}x^{-3/2}y^{1/2} + \frac{1}{2}x^{-1}\right) + \left(-x^{-1/2}y^{1/2}\right) - 1$\n$= x^{-1/2}y^{1/2} + 1 - x^{-1/2}y^{1/2} - 1$\n$= 0 = \text{RHS}$\nHence, $2x\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right) - 1 = 0$ is shown.