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The product rule is an essential tool in differentiation when dealing with products of two functions. In simple terms, if you have a function that is the product of two separate functions, say \( u(x) \) and \( v(x) \), then the product rule allows you to find the derivative of the entire product \( y(x) = u(x) \cdot v(x) \). The formula to find the derivative using the product rule is:

• \( (uv)' = u'v + uv' \)

Here, \( u'(x) \) is the derivative of \( u(x) \), and \( v'(x) \) is the derivative of \( v(x) \).

When applying this to our example function \( y = x^2 \log_2(3 - 2x) \), we identify:

• \( u = x^2 \)
• \( v = \log_2(3-2x) \)

The next step involves finding the derivative of each part separately before plugging them back into the product rule. This systematic approach helps prevent mistakes and ensures you understand how each part contributes to the whole function's derivative.

The chain rule is another vital rule in differentiation, used when you encounter composite functions. When a function is composed of another function, the chain rule helps you differentiate them effectively. The general formula for applying the chain rule is:

• \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \)

This means you first differentiate the outer function and then multiply by the derivative of the inner function.

In our exercise, dealing with the function \( v = \log_2(3 - 2x) \), we treat it as a composite function \( \log_2(g(x)) \) where \( g(x) = 3 - 2x \). By the chain rule, we differentiate like this:

• Find \( g'(x) = -2 \).
• Use the derivative formula for the logarithmic function which is \( \frac{1}{g(x)\ln(2)} \) to find \( v'(x) \).

Combining these, we get:

• \( v'(x) = \frac{-2}{(3 - 2x)\ln(2)} \)

Using the chain rule correctly ensures all parts of a composite function are accounted for in differentiation.

Logarithmic differentiation is a powerful technique specifically useful for functions where the base of a logarithm is not A familiar value like e. In such cases, you apply the rules of logarithmic derivatives to simplify the process. The steps involve utilizing the property of logarithms:

• \( \frac{d}{dx}\log_a(f(x)) = \frac{1}{f(x)\ln(a)} \cdot f'(x) \)

For our function \( \log_2(3 - 2x) \), the base is 2, which requires adjusting from the usual natural base \( e \) (where \( \ln \) is the natural logarithm) to base 2.

Therefore, to differentiate \( v = \log_2(3 - 2x) \), we followed these steps:

• First, recognize \( f(x) = 3 - 2x \) within the logarithm.
• Find \( f'(x) = -2 \).
• Apply the logarithmic differentiation formula to get \( v'(x) = \frac{-2}{(3 - 2x)\ln(2)} \).

This technique efficiently handles the complications arising from non-standard bases in logarithms.