### Lecture on the Product Rule for Derivatives Today, we'll explore the product rule for derivatives, an essential tool in calculus for finding the derivative of the product of two functions. To illustrate this, we'll consider the derivative of the product $x^2 \cdot x^3$. We'll approach this in two ways: first, by simplifying the expression and then using the power rule, and second, by directly applying the product rule. --- #### Simplifying Before Differentiation Let's start by simplifying $x^2 \cdot x^3$. According to the laws of exponents, when we multiply two expressions with the same base, we add their exponents: $$ x^2 \cdot x^3 = x^{2+3} = x^5 $$ --- Now, to find the derivative of $x^5$, we use the power rule. The power rule states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Applying this to $x^5$, we get: $$ \frac{d}{dx}x^5 = 5x^{5-1} = 5x^4 $$ Thus, by simplifying the expression first and then differentiating, we find that the derivative of $x^2 \cdot x^3$ is $5x^4$. --- #### Using the Product Rule Now, let's approach the same problem using the product rule. The product rule states that if we have two functions $u(x)$ and $v(x)$, and $y = u(x)v(x)$, then the derivative of $y$ with respect to $x$ is given by: $$ y' = u'v + uv' $$ --- In our case, $u(x) = x^2$ and $v(x) = x^3$. To apply the product rule, we first need to find the derivatives of $u(x)$ and $v(x)$. Using the power rule: - The derivative of $u(x) = x^2$ is $u'(x) = 2x^{2-1} = 2x$. - The derivative of $v(x) = x^3$ is $v'(x) = 3x^{3-1} = 3x^2$. --- Housing the information in this table as follows: | | | |--- | ---| | $u=x^2$ | $v=x^3$ | | $u'=2x$ | $v'=3x^2$ | --- Applying the product rule: \begin{align*} y' &= u'v + uv' \\ &=(2x)(x^3) + (x^2)(3x^2) \\ &=2x^4 + 3x^4 \\ &=5x^4 \end{align*} Interestingly, both methods yield the same result: $5x^4$ This not only demonstrates the consistency and reliability of calculus principles but also shows two different approaches to solving the same problem. --- #### Conclusion By comparing these two methods, we can appreciate the flexibility in choosing an approach based on the problem at hand. Simplifying before differentiating can often make the process quicker and more straightforward. However, the product rule is indispensable when dealing with more complex functions that cannot be easily simplified or when the functions represent quantities that are not easily combined. This illustrates the power and utility of calculus tools in mathematical analysis and problem-solving.