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The product rule is an essential tool in calculus used to differentiate functions that are multiplied together. When you have two functions, say \( u(t) \) and \( v(t) \), their product \( s(t) = u(t) \cdot v(t) \) can be differentiated using the product rule. It states that:

• \( s'(t) = u'(t)v(t) + u(t)v'(t) \)

This formula allows us to find the derivative of a product by simply differentiating each function separately and then combining the results as shown. In practical applications, this rule is particularly useful when working with complex polynomials and exponential functions, allowing you to break down the differentiation process into simpler steps. So remember, whenever you encounter a product of two functions, think of the product rule as your go-to method for differentiation.

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of a function concerning its variable. It's like calculating how fast or slow something changes at a particular point.
In our example, we need to find the derivatives of \( u(t) = t^{4/3} \) and \( v(t) = \frac{1}{e^{t}} \). To differentiate \( u(t) \), we apply the power rule, which states \( u'(t) = \frac{d}{dt} t^n = n t^{n-1} \). Thus, \( u'(t) = \frac{4}{3}t^{1/3} \).
For \( v(t) \), note it's an exponential decay function \( v(t) = e^{-t} \). Differentiating \( e^{-t} \) gives us \( v'(t) = -e^{-t} \). This result demonstrates how differentiation can handle both polynomial and exponential functions efficiently, setting the stage for more complex calculations.

Once you have applied the product rule and found the derivative, the next step is simplifying the expression. Simplification makes the derivative easier to interpret and use. Simplifying means combining like terms and reducing fractions or coefficients where possible.
In our exercise, after applying the product rule, we arrived at:

• \( s'(t) = \frac{4}{3}t^{1/3}e^{-t} - t^{4/3}e^{-t} \)

You can spot that both terms share a common factor: \( e^{-t} \). Factoring this out, if needed, can further streamline the expression. Simplification helps to readily see how the function behaves and how different parts of the expression interact with each other, particularly when interpreting or graphing the derivative.