If \(n\) is any real number and \(a\) is a constant, then:

Writing the derivative as \(f'(x)\) or as \(\dfrac{dy}{dx}\) means the same thing. We usually use \(f(x)\) when thinking of a function, and \(y\) when thinking of its graph.

Apply the rule “multiply by the power, then reduce the power by 1”.

• \[ f(x) = x^{2} \;\;\Rightarrow\;\; f'(x) = 2x^{2-1} = 2x \]
• \[ f(x) = 5x^{3} \;\;\Rightarrow\;\; f'(x) = 5 \times 3 x^{3-1} = 15x^{2} \]
• \[ y = -4x^{5} \;\;\Rightarrow\;\; \frac{dy}{dx} = -4 \times 5 x^{5-1} = -20x^{4} \]
• \[ y = \frac{2}{3}x^{7} \;\;\Rightarrow\;\; \frac{dy}{dx} = \frac{2}{3} \times 7 x^{7-1} = \frac{14}{3}x^{6} \]

The same rule works for negative and fractional powers. First rewrite roots or reciprocals as powers of \(x\).

• \[ f(x) = 2x^{-1} \;\;\Rightarrow\;\; f'(x) = 2 \times (-1)x^{-1-1} = -2x^{-2} \]
• \[ f(x) = 5\sqrt{x} = 5x^{1/2} \;\;\Rightarrow\;\; f'(x) = 5 \times \frac{1}{2} x^{\frac{1}{2}-1} = \frac{5}{2}x^{-1/2} = \frac{5}{2\sqrt{x}} \]
• \[ y = \frac{4}{5}x^{2/3} \;\;\Rightarrow\;\; \frac{dy}{dx} = \frac{4}{5} \times \frac{2}{3} x^{\frac{2}{3}-1} = \frac{8}{15}x^{-1/3} \]
• \[ y = \frac{1}{x^{4}} = x^{-4} \;\;\Rightarrow\;\; \frac{dy}{dx} = -4x^{-5} = -\frac{4}{x^{5}} \]
• In general: \[ \frac{1}{x^{p}} = x^{-p}. \]

If \(a\) is a constant, then:

• \[ f(x) = 3 \;\;\Rightarrow\;\; f'(x) = 0 \]
• \[ f(x) = -5x \;\;\Rightarrow\;\; f'(x) = -5 \]
• \[ y = \sqrt{7} \;\;\Rightarrow\;\; \frac{dy}{dx} = 0 \]
• \[ y = x \;\;\Rightarrow\;\; \frac{dy}{dx} = 1 \]

If an expression is a sum or difference of terms, differentiate each term separately.

In general, if \[ y = f(x) + g(x) + h(x) + \dots, \] then \[ \frac{dy}{dx} = f'(x) + g'(x) + h'(x) + \dots \]

This rule is used all the time with polynomials and many other functions in 9709.

Question

Let \[ f(x) = \frac{8}{q\sqrt{x}} + \sqrt{x}, \] where \(q\) is a real constant and \(x > 0\). Given that \(f'(3) = -\dfrac{\sqrt{3}}{18}\), find the value of \(q\).

Solution

Step 1: Rewrite using powers of \(x\).

\[ f(x) = \frac{8}{q} x^{-1/2} + x^{1/2}. \]

Step 2: Differentiate term by term.

\[ \begin{aligned} f'(x) &= \frac{8}{q} \cdot \left(-\frac{1}{2}\right) x^{-3/2} + \frac{1}{2} x^{-1/2} \\[4pt] &= -\frac{4}{q} x^{-3/2} + \frac{1}{2} x^{-1/2}. \end{aligned} \]

Step 3: Substitute \(x = 3\).

\[ \begin{aligned} f'(3) &= -\frac{4}{q} \cdot 3^{-3/2} + \frac{1}{2} \cdot 3^{-1/2} \\[4pt] &= -\frac{4}{q} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{2} \cdot \frac{1}{\sqrt{3}} \\[4pt] &= -\frac{4}{3q\sqrt{3}} + \frac{1}{2\sqrt{3}}. \end{aligned} \]

Step 4: Write as a single fraction.

Use the common denominator \(6q\sqrt{3}\): \[ \begin{aligned} \frac{1}{2\sqrt{3}} &= \frac{3q}{6q\sqrt{3}}, \\[4pt] -\frac{4}{3q\sqrt{3}} &= -\frac{8}{6q\sqrt{3}}. \end{aligned} \] So \[ f'(3) = \frac{3q - 8}{6q\sqrt{3}}. \]

Step 5: Use the given value of \(f'(3)\).

We are told that \[ f'(3) = -\frac{\sqrt{3}}{18}. \] Therefore \[ \frac{3q - 8}{6q\sqrt{3}} = -\frac{\sqrt{3}}{18}. \]

Step 6: Solve for \(q\).

Multiply both sides by \(18q\sqrt{3}\): \[ \begin{aligned} 18q\sqrt{3} \cdot \frac{3q - 8}{6q\sqrt{3}} &= 18q\sqrt{3} \cdot \left(-\frac{\sqrt{3}}{18}\right) \\[4pt] 3(3q - 8) &= -3q. \end{aligned} \] So \[ \begin{aligned} 9q - 24 &= -3q \\[4pt] 12q &= 24 \\[4pt] q &= 2. \end{aligned} \]

Answer: \(q = 2\).