Let's answer each question step by step: --- ### **Question 4**: **Dividing the rectangle** We start with a rectangle whose area is \((f(t+h)) \cdot (g(t+h))\). We want to break this area into four smaller rectangles, using the idea that \(f(t+h) = f(t) + \Delta f\) and \(g(t+h) = g(t) + \Delta g\), where \(\Delta f = f(t+h) - f(t)\) and \(\Delta g = g(t+h) - g(t)\). So, the rectangle can be written as: \[ f(t+h)g(t+h) = [f(t) + \Delta f][g(t) + \Delta g] \] Expanding this: \[ = f(t)g(t) + f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g \] So, the four parts are: - \(f(t)g(t)\): Area of the original rectangle at \(t\) - \(f(t)\Delta g\): Rectangle with width \(f(t)\) and height \(\Delta g\) - \(g(t)\Delta f\): Rectangle with width \(g(t)\) and height \(\Delta f\) - \(\Delta f \Delta g\): Smallest rectangle, area due to both \(f\) and \(g\) changing **Largest rectangle:** \(f(t)g(t)\) **Expression:** \[ f(t+h)g(t+h) = f(t)g(t) + f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g \] --- ### **Question 5**: **Limit definition and rectangles** Recall: \[ \frac{d}{dt}(f(t)g(t)) = \lim_{h \to 0} \frac{f(t+h)g(t+h) - f(t)g(t)}{h} \] From above, \(f(t+h)g(t+h) = f(t)g(t) + f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g\). So, \[ f(t+h)g(t+h) - f(t)g(t) = f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g \] So the numerator is the sum of the areas of the three rectangles **other than** the original one (the changes as you nudge both functions). **Interpretation:** - \(f(t)\Delta g\): Area due to change in \(g\) - \(g(t)\Delta f\): Area due to change in \(f\) - \(\Delta f \Delta g\): Area due to both changing **The limit:** As \(h \to 0\), \(\Delta f\) and \(\Delta g\) become infinitesimally small, and the last term \(\Delta f \Delta g\) becomes negligible compared to the other two. --- ### **Question 6**: **Connection to product rule** Plug into the limit: \[ \lim_{h \to 0} \frac{f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g}{h} \] Write \(\Delta f = f(t+h) - f(t), \Delta g = g(t+h) - g(t)\): \[ = \lim_{h \to 0} \left( f(t)\frac{\Delta g}{h} + g(t)\frac{\Delta f}{h} + \frac{\Delta f \Delta g}{h} \right) \] As \(h \to 0\): - \(\frac{\Delta f}{h} \to f'(t)\) - \(\frac{\Delta g}{h} \to g'(t)\) - \(\frac{\Delta f \Delta g}{h}\) goes to 0 (since both \(\Delta f, \Delta g \sim h\), so their product is \(\sim h^2\), and \(\frac{h^2}{h} \to 0\)) So, \[ \frac{d}{dt}(f(t)g(t)) = f(t)g'(t) + g(t)f'(t) \] **Conjecture for ignoring the last term:** The term \(\frac{\Delta f \Delta g}{h}\) can be ignored because as \(h \to 0\), it shrinks faster than the other terms (second-order smallness). This is why the product rule only contains two terms. --- ## **Final Answers:** ### **Q4:** \[ f(t+h)g(t+h) = f(t)g(t) + f(t)\Delta g + g(t)\Delta f + \Delta f \Delta g \] ### **Q5:** The numerator represents the combined area of the three rectangles reflecting the change due to \(f\), \(g\), and both. The limit shows how these changes behave as \(h \to 0\). ### **Q6:** The product rule arises because the extra term \(\frac{\Delta f \Delta g}{h}\) becomes negligible as \(h \to 0\). This term can be ignored since it's of higher order (goes to zero faster). Thus, \[ \frac{d}{dt}(f(t)g(t)) = f(t)g'(t) + g(t)f'(t) \]