Problem 24 - Precalculus: Mathematics for Calculus

Question

Find the derivative of the function at the given number. $$f(x)=\frac{x}{2-x}, \quad \text { at }-3$$

Step-by-Step Solution

Answer

The derivative at \( x = -3 \) is \( \frac{2}{25} \).

Step 1: Apply the Quotient Rule

The function given is \( f(x) = \frac{x}{2-x} \). To find its derivative, use the quotient rule for derivatives, which states: if \( h(x) = \frac{u(x)}{v(x)} \), then \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Here \( u(x) = x \) and \( v(x) = 2-x \).

Step 2: Differentiate \( u(x) \) and \( v(x) \)

Calculate the derivatives of the numerator and denominator functions: \( u(x) = x \) and \( v(x) = 2-x \). Therefore, \( u'(x) = 1 \) and \( v'(x) = -1 \).

Step 3: Apply the Derivatives in the Quotient Rule

Substitute \( u'(x) = 1 \), \( v(x) = 2-x \), \( u(x) = x \), and \( v'(x) = -1 \) into the quotient rule formula: \[ f'(x) = \frac{1 \cdot (2-x) - x \cdot (-1)}{(2-x)^2} \].

Step 4: Simplify the Resulting Expression

Simplify the expression: \[ f'(x) = \frac{2-x + x}{(2-x)^2} = \frac{2}{(2-x)^2}. \]

Step 5: Evaluate the Derivative at \( x = -3 \)

Substitute \( x = -3 \) into the derivative \( f'(x)=\frac{2}{(2-x)^2} \): \[ f'(-3) = \frac{2}{(2-(-3))^2} = \frac{2}{(2+3)^2} = \frac{2}{25}.\]

Key Concepts

Derivative Calculation

In calculus, finding the derivative of a function is one of the fundamental tasks. A derivative gives us an insight into the rate of change of a function with respect to a variable. In the context of the given exercise, the function we need to differentiate is a rational function, more specifically:

• \( f(x) = \frac{x}{2-x} \)

To find the derivative of such a rational function, we apply a rule known as the Quotient Rule. This formula is specifically designed for functions that are ratios of two simpler functions, denoted as \( u(x) \) and \( v(x) \), where:

• \( u(x) = x \)
• \( v(x) = 2-x \)

The rule states that the derivative of \( f(x) = \frac{u(x)}{v(x)} \) is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\]Breaking this down involves differentiating both the numerator and the denominator individually, then applying those derivatives in the formula. The simplicity of differentiation for linear functions helps here, as we easily find \( u'(x) = 1 \) and \( v'(x) = -1 \).

Simplifying Rational Expressions

After applying the Quotient Rule, it's crucial to simplify the resulting expression of the derivative to make it more manageable for further computations. Consider the result from applying the rule:\[f'(x) = \frac{1 \cdot (2-x) - x \cdot (-1)}{(2-x)^2}.\]To simplify, distribute and combine like terms:

• First, multiply: \(1 \times (2-x) = 2-x\)
• Then, we have: \(-x \times (-1) = x\)
• Combine the terms: \(2 - x + x = 2\)

Thus, the simplified form of the derivative is:\[f'(x) = \frac{2}{(2-x)^2}\]Simplifying rational expressions not only makes them look cleaner, but it also facilitates the evaluation process by reducing potential computational errors.

Evaluating Derivatives

Once the derivative is simplified, the next step is to evaluate it at a specific point, which tells us the rate of change of the function at that particular value of \( x \). In this exercise, the point of interest is \( x = -3 \). We substitute \( -3 \) into the simplified derivative:\[f'(x)=\frac{2}{(2-x)^2}\]Substituting \( x = -3 \) gives:\[f'(-3) = \frac{2}{(2-(-3))^2} = \frac{2}{(2+3)^2} = \frac{2}{25}\]By performing this substitution, we find that the rate of change of the function \( f(x) \) at \( x = -3 \) is \( \frac{2}{25} \). Evaluating derivatives is crucial in many practical applications, including physics and engineering, as it provides specific information about the behavior of a function at particular points.