复习题 2020/Promo/SAJC/Q3

1. Apply the product rule to the given expression.

Let $y = \text{e}^{2x} \cos^{-1}(3x)$.
Using the product rule $\frac{\text{d}}{\text{d}x}(uv) = v\frac{\text{d}u}{\text{d}x} + u\frac{\text{d}v}{\text{d}x}$, we set $u = \text{e}^{2x}$ and $v = \cos^{-1}(3x)$.

2. Differentiate each part with respect to $x$.

$\frac{\text{d}u}{\text{d}x} = 2\text{e}^{2x}$
$\frac{\text{d}v}{\text{d}x} = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot \frac{\text{d}}{\text{d}x}(3x) = -\frac{3}{\sqrt{1 - 9x^2}}$

3. Combine the results into the final expression.

$\frac{\text{d}y}{\text{d}x} = 2\text{e}^{2x} \cos^{-1}(3x) + \text{e}^{2x} \left( -\frac{3}{\sqrt{1 - 9x^2}} \right)$
$\frac{\text{d}y}{\text{d}x} = \text{e}^{2x} \left( 2\cos^{-1}(3x) - \frac{3}{\sqrt{1 - 9x^2}} \right)$

1. Differentiate the given equation implicitly with respect to $x$ to find an expression for $\frac{\text{d}y}{\text{d}x}$.

$x^{1/2} + y^{1/2} = a^{1/2}$
$\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2}\frac{\text{d}y}{\text{d}x} = 0$
$y^{-1/2}\frac{\text{d}y}{\text{d}x} = -x^{-1/2}$
$\frac{\text{d}y}{\text{d}x} = -x^{-1/2}y^{1/2}$

2. Differentiate $\frac{\text{d}y}{\text{d}x}$ with respect to $x$ using the product rule and implicit differentiation.

$\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left(-x^{-1/2}y^{1/2}\right)$
$= -\left[ \left(-\frac{1}{2}x^{-3/2}\right)y^{1/2} + x^{-1/2}\left(\frac{1}{2}y^{-1/2}\frac{\text{d}y}{\text{d}x}\right) \right]$
$= \frac{1}{2}x^{-3/2}y^{1/2} - \frac{1}{2}x^{-1/2}y^{-1/2}\frac{\text{d}y}{\text{d}x}$

3. Substitute the expression for $\frac{\text{d}y}{\text{d}x}$ into the second derivative and simplify.

Substitute $\frac{\text{d}y}{\text{d}x} = -x^{-1/2}y^{1/2}$ into the equation from step 2:
$\frac{\text{d}^2y}{\text{d}x^2} = \frac{1}{2}x^{-3/2}y^{1/2} - \frac{1}{2}x^{-1/2}y^{-1/2}\left(-x^{-1/2}y^{1/2}\right)$
$\frac{\text{d}^2y}{\text{d}x^2} = \frac{1}{2}x^{-3/2}y^{1/2} + \frac{1}{2}x^{-1}$

4. Substitute $\frac{\text{d}^2y}{\text{d}x^2}$ and $\frac{\text{d}y}{\text{d}x}$ into the target expression to verify it equals zero.

Evaluate the Left Hand Side (LHS) of the required equation:
$\text{LHS} = 2x\frac{\text{d}^2y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} - 1$
$= 2x\left( \frac{1}{2}x^{-3/2}y^{1/2} + \frac{1}{2}x^{-1} \right) + \left( -x^{-1/2}y^{1/2} \right) - 1$
$= \left(x^{-1/2}y^{1/2} + 1\right) - x^{-1/2}y^{1/2} - 1$
$= 0 = \text{RHS (shown)}$