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The Quotient Rule is a fundamental technique used in calculus for differentiating functions that are expressed as a ratio of two other functions. This is especially useful when you come across any function expressed in the form of a fraction, like our given function \( f(x) = \frac{x}{x+5} \). Let's break it down further to better understand how the Quotient Rule applies:

• Consider a function \( g(x) = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are both differentiable functions.
• The Quotient Rule states that the derivative \( g'(x) \) is defined as: \[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

This means, when applying the Quotient Rule, you multiply the derivative of the numerator by the denominator, subtract the product of the numerator and the derivative of the denominator, and then divide everything by the square of the denominator.
When we apply this to our function \( f(x) = \frac{x}{x+5} \):

• \( u(x) = x \) and \( v(x) = x+5 \).
• The derivatives are: \( u'(x) = 1 \) and \( v'(x) = 1 \).

Substituting into the Quotient Rule gives us the derivative \( f'(x) = \frac{1 \cdot (x+5) - x \cdot 1}{(x+5)^2} = \frac{5}{(x+5)^2} \).
This calculation allows us to understand the rate of change of the original function, considering its structure as a fraction.

Finding the derivative of an inverse function is slightly more intricate than finding the derivative of a regular function. It's crucial because it provides insights into how the inverse function behaves and changes over its domain. For a given function \( f \), if \( f(x) \) is its inverse denoted as \( f^{-1}(x) \), the derivative \( \left(f^{-1}(x)\right)' \) can be determined using:

• \[ \left( f^{-1}(x) \right)' = \frac{1}{f'(f^{-1}(x))} \]

This formula gives us a way to differentiate the inverse by relying on the derivative of the original function. By evaluating \( f'(f^{-1}(x)) \), you literally "plug in" the inverse function into the derivative of the original function. It's like seeing how the transformation through the inverse impacts the changes quantified by the derivative. This method helps take a more holistic view of how the function and its inverse are related through differentiation.
In our specific example, using the derivative \( f'(x) = \frac{5}{(x+5)^2} \) and plugging in \( f^{-1}(x) = \frac{-5x}{x-1} \), we calculate the derivative as follows: \( \left(f^{-1}(x)\right)' = \frac{1}{\frac{5}{((f^{-1}(x)+5)^2}} \), leading us to \( \frac{(x-1)^2}{25} \).
This value tells us how the inverse function \( f^{-1}(x) \) is changing in relation to \( x \).

When dealing with calculus and inverse functions, the derivative of an inverse function is a key concept that ties together how a function and its inverse relate through differentiation. The process is not just about flipping roles but also about understanding the relationship of their rates of change. The derivative of an inverse function can be found by using the formula:

• \[ \left( f^{-1}(x) \right)' = \frac{1}{f'(f^{-1}(x))} \]

This highlights the interconnectedness of the original and inverse functions through their derivatives. What you're doing here not only takes into consideration how \( f \) changes but also how \( f^{-1} \) steers the changes in \( x \). In other words, the behavior of the inverse is just the reciprocal of the rate of change of the function itself evaluated at the inverse.

For our function \( f(x) = \frac{x}{x+5} \) with \( f^{-1}(x) = \frac{-5x}{x-1} \), after substituting and simplifying, we've calculated \( \left( f^{-1}(x) \right)' = \frac{(x-1)^2}{25} \). This outcome wraps up how inverses make the relationship between \( y \) and \( x \) equally impactful, displaying the intricacy of calculus when functions and their inverses mingle.