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When you have a function that is the quotient of two other functions, you use the quotient rule to find its derivative. The quotient rule can be remembered with the saying: "Down D Up minus Up D Down, all over the square of down." In mathematical terms, if you have a function of the form \( h(x) = \frac{f(x)}{g(x)} \), the derivative is given by: \[ h'(x) = \frac{f'(x) g(x) - f(x) g'(x)}{[g(x)]^2} \] This means you take the derivative of the numerator times the denominator, subtract the numerator times the derivative of the denominator, and place it all over the denominator squared. In the context of our problem: - The "up" or numerator is \( u(x) = x f(x) \) - The "down" or denominator is \( v(x) = x + g(x) \) Applying the quotient rule requires knowing the derivatives of both \( u(x) \) and \( v(x) \).

The product rule is applied when differentiating a product of two functions. If \( u(x) = a(x) \cdot b(x) \), then according to the product rule, the derivative \( u'(x) \) is: \[ u'(x) = a'(x) b(x) + a(x) b'(x) \] This rule means you derive the first function while leaving the second unchanged, then add the unchanged first function multiplied by the derivative of the second function. In the exercise, the function \( u(x) \) is \( x f(x) \). So for its derivative: - The derivative of \( x \) is 1. - The derivative of \( f(x) \) is \( f'(x) \) Hence, \( u'(x) = 1 \cdot f(x) + x \cdot f'(x) = f(x) + x f'(x) \).

Differentiability is a fundamental concept in calculus indicating a function has a derivative at a specific point. A function is differentiable at a given point if there's a well-defined tangent line or derivative. Differentiable functions are inherently smooth, which means no sharp corners or cusps at the point of differentiation. In the exercise, both functions \( f(x) \) and \( g(x) \) are given as differentiable at \( x = 1 \). This is crucial because it allows the application of derivative rules like the quotient and product rules. The exercise differentiability ensures that we can safely use the function values \( f(1) \), \( g(1) \), and their derivatives \( f'(1) \), \( g'(1) \) in calculations.

The sum rule in calculus is simple and aids in finding the derivative of a sum of functions. It states: The derivative of a sum is the sum of the derivatives. Suppose \( v(x) = c(x) + d(x) \), its derivative is: \[ v'(x) = c'(x) + d'(x) \] In the exercise, the function \( v(x) \) is \( x + g(x) \). Applying the sum rule yields: - The derivative of \( x \) is 1. - The derivative of \( g(x) \) is \( g'(x) \) Therefore, \( v'(x) = 1 + g'(x) \). Simple and straightforward, it allows you to combine derivatives of simpler components into one.