Given that $y = \dfrac{4x}{\sqrt{3-2x}}$, show that $\dfrac{dy}{dx} = \dfrac{12-4x}{\sqrt{(3-2x)^3}}$. Hence find the value of $\displaystyle \int_{0}^{1} \dfrac{3 - x}{\sqrt{(3 - 2x)^3}}\, dx$.

1. Differentiate the numerator and denominator

Let $u = 4x$ and $v = (3-2x)^{\frac{1}{2}}$.
Then $\frac{du}{dx} = 4$.
Using the chain rule, $\frac{dv}{dx} = \frac{1}{2}(3-2x)^{-\frac{1}{2}}(-2) = -(3-2x)^{-\frac{1}{2}}$.

2. Apply the quotient rule

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{4(3-2x)^{\frac{1}{2}} - (4x)\left(-(3-2x)^{-\frac{1}{2}}\right)}{\left((3-2x)^{\frac{1}{2}}\right)^2}$

3. Simplify the numerator by factoring

$\frac{dy}{dx} = \frac{4(3-2x)^{\frac{1}{2}} + 4x(3-2x)^{-\frac{1}{2}}}{3-2x}$
Factor out $(3-2x)^{-\frac{1}{2}}$ from the numerator:
$\frac{dy}{dx} = \frac{(3-2x)^{-\frac{1}{2}} \left[ 4(3-2x)^1 + 4x \right]}{3-2x}$

4. Simplify to obtain the final required expression

$\frac{dy}{dx} = \frac{12 - 8x + 4x}{(3-2x)(3-2x)^{\frac{1}{2}}}$
$\frac{dy}{dx} = \frac{12 - 4x}{(3-2x)^{\frac{3}{2}}}$
$\frac{dy}{dx} = \frac{12 - 4x}{\sqrt{(3-2x)^3}}$ (Shown)

1. Relate the given integrand to the derivative found in part (a)

From (a), we have $\frac{d}{dx} \left( \frac{4x}{\sqrt{3-2x}} \right) = \frac{12-4x}{\sqrt{(3-2x)^3}}$.
Notice that $\frac{12-4x}{\sqrt{(3-2x)^3}} = 4 \left( \frac{3-x}{\sqrt{(3-2x)^3}} \right)$.

2. Express the integral using the reverse of differentiation

$\int \frac{3-x}{\sqrt{(3-2x)^3}} \, dx = \frac{1}{4} \int \frac{12-4x}{\sqrt{(3-2x)^3}} \, dx$
$= \frac{1}{4} \left( \frac{4x}{\sqrt{3-2x}} \right) = \frac{x}{\sqrt{3-2x}}$

3. Evaluate the definite integral using the limits

$\int_{0}^{1} \frac{3 - x}{\sqrt{(3 - 2x)^3}} \, dx = \left[ \frac{x}{\sqrt{3-2x}} \right]_{0}^{1}$
$= \frac{1}{\sqrt{3-2(1)}} - \frac{0}{\sqrt{3-2(0)}}$
$= \frac{1}{\sqrt{1}} - 0 = 1$