91Ó°ÊÓ

In calculus, when you're dealing with the product of two functions, it's important to use the product rule to find the derivative. This rule is essential when you have a product of functions like in the exercise: a function multiplied by another, such as \( y = f(t) \cdot g(t) \). The product rule is a formula that allows you to find the derivative of this product, where the derivative of the product \( (fg)' \) is given by:

• \( (fg)' = f'g + fg' \)

This means you take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function. Each component needs to be differentiated appropriately, ensuring that all parts of the expression contribute to the final derivative result.

The chain rule is a fundamental technique used in calculus to differentiate composite functions. These are functions where one function is nested inside another, like \( g(t) = (2t^2 - 5)^4 \) in the exercise. The chain rule helps us find the derivative of such a function by breaking it down into simpler components. Using the chain rule involves a few steps:

• Identify the inner function, which in this case is \( u = 2t^2 - 5 \).
• Identify the outer function, represented as \( u^4 \).
• Differentiate the outer function with respect to the inner function: \( 4u^3 \).
• Differentiate the inner function \( u \) with respect to \( t \): \( \frac{du}{dt} = 4t \).

By multiplying these derivatives together, using \( g'(t) = \frac{dg}{du} \cdot \frac{du}{dt} \), we can find the derivative of the composite function. This method ensures that each layer of the function is correctly differentiated, maintaining the integrity of the original equation.

When it comes to finding the derivative of polynomial functions, the process involves applying basic rules of differentiation, such as the power rule. Polynomials are expressions made of variables raised to non-negative integer powers, and possibly multiplied by coefficients. For example, a simple polynomial might look like \(f(t) = 3t \). To differentiate this, we apply the rule:

• If \( f(t) = at^n \), then \( f'(t) = an\cdot t^{n-1} \).

In the exercise provided, the function \( f(t) = 3t \) translates to a straightforward derivative \( f'(t) = 3 \), as it is a linear term. When seeking to simplify expressions or solve for derivatives, recognizing and applying these fundamental rules correctly is crucial. They provide a foundation for more complex differentiation tasks, where polynomials often appear as parts of larger functions.