A function $f$ is defined on a suitable domain by $$f(x) = -rac{ ext{sqrt}(3x - 2)}{x ext{sqrt}(x)}.$$ Find $f'(4)$. - Scottish Highers Maths - Question 7 - 2015

We will use the quotient rule, which states that if we have a function in the form of $\frac{u}{v}$, then:

$\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.$

Here, let:

• $u = (3x - 2)^{1/2}$, so $u' = \frac{3}{2(3x - 2)^{1/2}}$
• $v = x^{3/2}$, so $v' = \frac{3}{2}x^{1/2}$.

Applying the rule:

$f'(x) = -\frac{\frac{3}{2}(3x - 2)^{1/2} \cdot x^{3/2} - (3x - 2)^{1/2} \cdot \frac{3}{2}x^{1/2}}{(x^{3/2})^2}.$

Let's simplify the expression we derived. Rearranging will help us get:

$f'(x) = -\frac{\frac{3}{2}(3x - 2)^{1/2} x^{3/2} - \frac{3}{2}(3x - 2)^{1/2} x^{1/2}}{x^3}.$

Factoring out common terms yields:

$f'(x) = -\frac{\frac{3}{2}(3x - 2)^{1/2}}{x^3}\left(x - 1\right).$

Substituting $x = 4$ into our derivative expression:

$f'(4) = -\frac{\frac{3}{2}(3(4) - 2)^{1/2}}{4^3}\left(4 - 1\right).$

Calculating further:

$f'(4) = -\frac{\frac{3}{2}(10)^{1/2}}{64}\cdot 3 = -\frac{3\sqrt{10}}{128}.$

Thus, the final answer is:

$f'(4) = -\frac{3\sqrt{10}}{128}.$