10. Demonstrate and explain how to find the derivative of the following functions. Be sure to explicitly denote which derivative rules (scalar multiple, sum/difference, quotient, etc.) you are using in your work. \( \begin{array}{l}\text { (g) } \\ f(w)=\frac{6 w^{2}-w-3}{w^{3}}\end{array} \)

We are given   f(w) = (6w² – w – 3) / w³. One approach is to first rewrite f(w) as a sum of individual terms and then differentiate term‐by‐term using the power rule. Alternatively, you can use the quotient rule. We’ll show both methods. ────────────────────────────── Method 1: Rewriting the Expression Step 1. Rewrite f(w) by dividing each term in the numerator by w³:   f(w) = (6w²)/w³ – (w)/w³ – 3/w³     = 6w^(2–3) – w^(1–3) – 3w^(–3)     = 6w^(–1) – w^(–2) – 3w^(–3). Step 2. Differentiate f(w) term-by-term using the power rule. Recall that if f(w) = w^n, then f '(w)= n w^(n–1), and also remember the scalar multiple rule. • For 6w^(–1):   d/dw [6w^(–1)] = 6 · (–1) w^(–2) = –6w^(–2). • For –w^(–2):   d/dw [–w^(–2)] = – (–2) w^(–3) = 2w^(–3). • For –3w^(–3):   d/dw [–3w^(–3)] = –3 · (–3) w^(–4) = 9w^(–4). Step 3. Combine the results:   f '(w) = –6w^(–2) + 2w^(–3) + 9w^(–4). This can also be written as:   f '(w) = –6/w² + 2/w³ + 9/w⁴ or, combining into a single rational expression, multiply numerator and denominator by a common factor (here, w⁴):   f '(w) = (–6w² + 2w + 9) / w⁴. ────────────────────────────── Method 2: Using the Quotient Rule The quotient rule states that if f(w) = g(w)/h(w), then   f '(w) = [g '(w)h(w) – g(w)h '(w)] / [h(w)]². Let   g(w) = 6w² – w – 3  and  h(w) = w³. First, find the derivatives of g and h: • g '(w) = d/dw [6w² – w – 3] = 12w – 1. • h '(w) = d/dw [w³] = 3w². Now apply the quotient rule:   f '(w) = [(12w – 1) · w³ – (6w² – w – 3) · (3w²)] / (w³)². Simplify step-by-step: 1. Multiply out (12w – 1) · w³:   = 12w⁴ – w³. 2. Multiply out (6w² – w – 3) · (3w²):   = 18w⁴ – 3w³ – 9w². 3. Substitute back:   f '(w) = [12w⁴ – w³ – (18w⁴ – 3w³ – 9w²)] / w⁶.       = [12w⁴ – w³ – 18w⁴ + 3w³ + 9w²] / w⁶.       = [–6w⁴ + 2w³ + 9w²] / w⁶. 4. Factor w² from the numerator:   = w²(–6w² + 2w + 9) / w⁶. 5. Cancel w² with part of w⁶:   = (–6w² + 2w + 9) / w⁴. This is the same result as in Method 1. ────────────────────────────── Final Answer The derivative of f(w) is:   f '(w) = –6w^(–2) + 2w^(–3) + 9w^(–4) or equivalently,   f '(w) = (–6w² + 2w + 9) / w⁴. In our work we used:  • The quotient rule (in Method 2);  • Rewriting into power form and the scalar multiple rule and the power rule (in Method 1);  • And the sum/difference rule in differentiating term-by-term. This completes the demonstration and explanation.