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The given function is:

\[
f(x) = 2 - \frac{x^2}{3} + \ln\left(\frac{x}{4}\right), \text{ where } x > 0
\]

Let's go through the parts of the question one by one:

### (a) Find an expression for \( f'(x) \).
We need to differentiate \( f(x) \) with respect to \( x \).

1. \( \frac{d}{dx} \left( 2 \right) = 0 \)
2. \( \frac{d}{dx} \left( -\frac{x^2}{3} \right) = -\frac{2x}{3} \)
3. \( \frac{d}{dx} \left( \ln \left( \frac{x}{4} \right) \right) \) can be simplified using the chain rule:
   \[
   \frac{d}{dx} \left( \ln \left( \frac{x}{4} \right) \right) = \frac{d}{dx} \left( \ln(x) - \ln(4) \right) = \frac{1}{x}
   \]

Therefore, the derivative \( f'(x) \) is:

\[
f'(x) = -\frac{2x}{3} + \frac{1}{x}
\]

---

### (b) Find the value of \( x \) in exact surd form if \( f'(x) = 0 \).

Set \( f'(x) = 0 \):

\[
-\frac{2x}{3} + \frac{1}{x} = 0
\]

Multiply through by \( x \) to eliminate the fraction:

\[
-\frac{2x^2}{3} + 1 = 0
\]

Rearrange:

\[
\frac{2x^2}{3} = 1 \quad \Rightarrow \quad x^2 = \frac{3}{2}
\]

Thus,

\[
x = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}
\]

---

### (c) Hence, find the stationary point of \( f(x) \) in exact surd form.

The stationary point occurs where \( f'(x) = 0 \). From part (b), we found \( x = \frac{\sqrt{6}}{2} \). To find the corresponding \( y \)-coordinate, substitute this value of \( x \) back into \( f(x) \):

\[
f\left( \frac{\sqrt{6}}{2} \right) = 2 - \frac{\left( \frac{\sqrt{6}}{2} \right)^2}{3} + \ln \left( \frac{\frac{\sqrt{6}}{2}}{4} \right)
\]

Simplify each term:

1. \( \left( \frac{\sqrt{6}}{2} \right)^2 = \frac{6}{4} = \frac{3}{2} \), so:
   \[
   \frac{\frac{3}{2}}{3} = \frac{1}{2}
   \]
   Thus, the second term becomes \( -\frac{1}{2} \).
   
2. \( \frac{\frac{\sqrt{6}}{2}}{4} = \frac{\sqrt{6}}{8} \), so the logarithmic term becomes:
   \[
   \ln\left( \frac{\sqrt{6}}{8} \right) = \ln(\sqrt{6}) - \ln(8) = \frac{1}{2} \ln(6) - \ln(8)
   \]

Thus, the exact value of \( f\left( \frac{\sqrt{6}}{2} \right) \) is:

\[
f\left( \frac{\sqrt{6}}{2} \right) = 2 - \frac{1}{2} + \frac{1}{2} \ln(6) - \ln(8)
\]

---

### (d) At the point \( x = \sqrt{3} \), find the gradient of \( f(x) \).

To find the gradient at \( x = \sqrt{3} \), substitute \( x = \sqrt{3} \) into \( f'(x) \):

\[
f'(\sqrt{3}) = -\frac{2\sqrt{3}}{3} + \frac{1}{\sqrt{3}} = -\frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = -\frac{1}{\sqrt{3}}
\]

---

Would you like more details on any specific step or explanation? 

Here are five related questions to expand on this material:

1. How do you find the second derivative of \( f(x) \) and interpret its significance?
2. What is the physical meaning of the stationary point in terms of optimization?
3. How would the presence of a logarithmic term impact the long-term behavior of \( f(x) \) as \( x \to \infty \)?
4. Could you calculate the concavity of the function at \( x = \frac{\sqrt{6}}{2} \)?
5. What would be the domain and range of the given function based on the constraints?

**Tip:** When dealing with logarithmic terms, it's helpful to simplify them using properties such as \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).

The function f is defined as f(x) = 2 - (x^2)/3 + ln(x/4), x > 0. a) Find an expression for f'(x). b) Find the value of x in exact surd form if f'(x) = 0. c) Hence, find the stationary point of f(x) in exact surd form. d) At the point x = sqrt(3), find the gradient of f(x).

Learn how to differentiate a function involving quadratic and logarithmic terms, find stationary points in exact surd form, and calculate gradients at specific points. This detailed step-by-step solution is ideal for students studying calculus at the Grade 12 or first-year university level.

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