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When you encounter the derivative of a function that's a quotient of two other functions, turning to the quotient rule is essential. The quotient rule helps you differentiate a function in the form \( \frac{u(x)}{v(x)} \). It's like you're taking apart the fraction to see how each piece changes. The rule itself can be remembered with this formula:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

What happens in this formula? You find the derivative of the top (\(u(x)\)), and the bottom (\(v(x)\)), then use them in the formula. You multiply the derivative of the top with the bottom function, subtract the product of the top function and the derivative of the bottom, and divide it all by the bottom squared.
This ensures you're considering every aspect of the chain reaction in both the numerator and denominator.

Exponential functions appear a lot, often looking like \( e^x \). They have a unique quality that makes them quite special: they grow really fast, and their derivative is exactly the same as the original function!

• \( \frac{d}{dx}(e^x) = e^x \)

This means the slope of an exponential function at any point is the value of the function itself at that point. That's why, in our problem, finding the derivative of \(e^x\) is straightforward. It stays \(e^x\), every time. It’s like magic, but definitely within the rules of calculus.
Remembering this property can save you a lot of time during differentiation as it removes one step entirely.

After applying the quotient rule, you might find yourself with a complex expression. Simplifying this expression is crucial for making it understandable and usable.

• Start by factoring out common terms in the numerator and denominator.
• Look out for terms that can cancel each other.

In the given exercise, once the quotient rule is applied, you end up with \( f'(x) = \frac{e^x x^2 - e^x 2x}{x^4} \). Notice how \(e^x\) is common in the numerator, allowing you to factor it out, which gives us \( f'(x) = \frac{e^x(x^2 - 2x)}{x^4} \).
Further simplification leads to cancelling \( x \) from both the numerator and denominator, resulting in the final form: \( f'(x) = \frac{e^x (x - 2)}{x^3} \). Clearing out terms where possible makes evaluating or further manipulating the function much easier. Each simplification step ensures a clearer path to understand how the function changes, making the derivative functional and neat.