The image shows a step-by-step proof of the product rule for differentiation:

$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$

This is derived using the definition of the derivative as a limit, specifically:

$\frac{d}{dx}(f(x)g(x)) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$

Step-by-step Explanation:

1. Start with the definition of the derivative for the product $f(x)g(x)$: $\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$

2. Split the expression inside the limit by adding and subtracting $f(x+h)g(x)$: $= \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}$ This step allows you to separate terms that involve $f(x)$ and $g(x)$.

3. Factor out $f(x+h)$ from the first two terms and $g(x)$ from the last two terms: $= \lim_{h \to 0} \left[ \frac{f(x+h)(g(x+h) - g(x))}{h} + \frac{g(x)(f(x+h) - f(x))}{h} \right]$

4. Now apply the limit to each term individually:

   • For the first term, $\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = g'(x)$, so the first term becomes $f(x)g'(x)$.
   • For the second term, $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x)$, so the second term becomes $g(x)f'(x)$.

5. Thus, you get: $f'(x)g(x) + f(x)g'(x)$ which is the product rule for derivatives.

Would you like further details or have any questions?

Here are some related questions to expand on this topic:

1. How can the product rule be applied to higher-order derivatives?
2. What is the significance of the product rule in practical applications of calculus?
3. How does the product rule relate to the chain rule in calculus?
4. Can you provide an example of using the product rule in a real-world problem?
5. What are some common mistakes students make when applying the product rule?

Tip: The product rule is fundamental when differentiating functions that are products of two or more variables, so practice applying it to different types of functions.