91Ó°ÊÓ

In calculus, the product rule is used when you want to find the derivative of a product of two functions. Instead of directly trying to differentiate a combined expression, you use each function's individual derivatives to get the final result. This makes your calculations easier and more organized.

The product rule formula can be written as:

• \( (uv)' = u'v + uv' \)

Here, \( u \) and \( v \) are functions of \( x \). The product rule states that the derivative of the product of \( u \) and \( v \) is equal to the derivative of \( u \) times \( v \), plus \( u \) times the derivative of \( v \).

To make this concept clearer, let's consider the function from the original exercise:

• \( u(x) = 8x^2 + 13 \)
• \( v(x) = \frac{39}{1 + 15e^{-0.09x}} \)

Use the product rule to find the derivative, \( f'(x) \). Calculate \( u'(x) \) and \( v'(x) \) first, then substitute them into the product rule formula.

The chain rule is essential when dealing with compositions of functions, meaning one function is inside another. It helps us take derivatives step by step by following a simple process: differentiate the outside function, then multiply by the derivative of the inside function.

For the given \( v(x) = \frac{39}{1 + 15e^{-0.09x}} \), notice that it's a composition where \( v(x) = \frac{39}{w(x)} \), with \( w(x) = 1 + 15e^{-0.09x} \). Apply the chain rule as follows:

• Find \( w'(x) \), the derivative of \( w(x) \) which is \( 1.35e^{-0.09x} \).
• The chain rule gives \( v'(x) = -\frac{39}{(w(x))^2} \cdot w'(x) \).
• Substitute \( w'(x) \) to get the full derivative of \( v(x) \): \(-\frac{39 \times 1.35e^{-0.09x}}{(1 + 15e^{-0.09x})^2} \).

This process highlights how the chain rule simplifies working with layered functions by breaking them down into manageable parts.

The power rule is one of the simplest and most commonly used rules in calculus for finding derivatives. It applies when you have a function of the form \( x^n \), where \( n \) is a constant. According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \).

In the given exercise, to differentiate \( u(x) = 8x^2 + 13 \), apply the power rule:

• The derivative of \( 8x^2 \) is \( 16x \).
• The constant \( 13 \) does not change when differentiated, so its derivative is \( 0 \).

Thus, \( u'(x) \) becomes \( 16x \).

The power rule provides a quick and easy way to derive polynomials, reflecting how even simple rules can achieve significant results when calculating derivatives.