91Ó°ÊÓ

Calculating the derivative of a function is crucial in understanding how the function behaves and changes at any given point. In this exercise, the quotient rule is used since the function is expressed as a fraction where one function is divided by another. The quotient rule provides a structured way to handle derivatives of such fractions.
The rule is expressed as: if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is:

• \( \frac{u'(x)v(x) - v'(x)u(x)}{(v(x))^2} \)

Here, \( u(x) \) and \( v(x) \) are the main functions in the numerator and denominator, respectively. This formula allows us to account for how changes in both parts (numerator and denominator) affect the overall rate of change of \( f(x) \). The derivative tells us the rate at which \( f(x) \) is changing as \( x \) increases.

The first step in using the quotient rule is to find the derivative of the numerator, \( u(x) = 3 - 2e^x \). Differentiation of a constant like 3 leads to 0, while differentiating \(-2e^x\) remains \(-2e^x\) since the derivative of an exponential function \( e^x \) is \( e^x \) itself. Consequently, the derivative of the numerator is:

• \( u'(x) = -2e^x \)

This derivative reflects how quickly the numerator of the function changes with respect to \( x \). The presence of \( -2e^x \) in the derivative indicates that the numerator's decrease is directly tied to the exponential rate, dependent on the value of \( x \). This understanding is essential when substituting back into the quotient rule formula.

Finding the derivative of the denominator, \( v(x) = 1 - 2x \), is the next step. Differentiating \( 1 \) yields \( 0 \). For the term \( -2x \), the derivative is simply \( -2 \), as it involves a linear function where the coefficient is carried through. So, the derivative here is:

• \( v'(x) = -2 \)

Knowing this helps us assess how the rate of change in the denominator affects the overall fraction. Often, the denominator's behavior directly influences the variations of the entire function, especially its zeros and any asymptotic behavior. Reducing confused terms in a function's derivative is possible through careful differentiation of the denominator.

Once the derivatives of the numerator and denominator are found, they are substituted into the quotient rule formula. This substitution gives us a complex expression that often needs simplification to make it interpretable and useful. After substitution, we have:

• \( f'(x) = \frac{(-2e^x)(1 - 2x) - (-2)(3 - 2e^x)}{(1-2x)^2} \)

Expanding the brackets and combining like terms helps reduce the expression to a more digestible form. Specifically, the steps lead to:

• Expanding: \((-2e^x + 4xe^x) + (6 - 4e^x) \)
• Combining: \( 4xe^x - 6e^x + 6 \)

The resulting simplified form sheds light on how different terms in the function contribute to its overall change. Having a simplified version ensures that the expression is not only easier to evaluate but also sets the stage for further analysis, such as sketching the derivative or understanding its pattern over different \( x \) values.