Problem 129 - Calculus Volume 1

Question

Assume that \(f(x)\) and \(g(x)\) are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives. $$ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 3 & 5 & -2 & 0 \\ \hline \boldsymbol{g}(\boldsymbol{x}) & 2 & 3 & -4 & 6 \\ \hline \boldsymbol{f}^{\prime}(\boldsymbol{x}) & -1 & 7 & 8 & -3 \\ \hline \boldsymbol{g}^{\prime}(\boldsymbol{x}) & 4 & 1 & 2 & 9 \\ \hline \end{array} $$ Find \(h^{\prime}(4)\) if \(h(x)=\frac{1}{x}+\frac{g(x)}{f(x)}\).

Step-by-Step Solution

Answer

The derivative \( h'(4) \) is undefined.

Step 1: Apply the Quotient Rule

For the function \( h(x) = \frac{1}{x} + \frac{g(x)}{f(x)} \), the derivative \( h'(x) \) involves the differentiation of \( \frac{g(x)}{f(x)} \). We apply the quotient rule: \[\left( \frac{g(x)}{f(x)} \right)' = \frac{f(x)g'(x) - g(x)f'(x)}{(f(x))^2}.\]

Step 2: Differentiate \( \frac{1}{x} \)

The derivative of \( \frac{1}{x} \) with respect to \( x \) is \( -\frac{1}{x^2} \).

Step 3: Combine the Derivatives

Combine the results of Step 1 and Step 2 to find \( h'(x) \):\[h'(x) = -\frac{1}{x^2} + \frac{f(x)g'(x) - g(x)f'(x)}{(f(x))^2}.\]

Step 4: Substitute Values at \( x = 4 \)

Using the table, substitute \( f(4) = 0 \), \( g(4) = 6 \), \( f'(4) = -3 \), and \( g'(4) = 9 \) into the expression:\[h'(4) = -\frac{1}{16} + \frac{0 \times 9 - 6 \times (-3)}{0^2}.\]

Step 5: Calculate and Analyze

Calculating \( h'(4) \) gives a division by zero in \( \frac{0 \times 9 - 6 \times (-3)}{0^2} \). This indicates that the derivative is undefined due to the division by zero in the denominator and numerator separately balancing each other out. Thus, \( h'(4) \) is undefined.

Key Concepts

Quotient Rule: Mastering Derivatives of Fractions

The "quotient rule" is a powerful tool for differentiating fractions. Specifically, it's used when you need to find the derivative of a ratio of two functions. Imagine you have two differentiable functions, say, \( g(x) \) and \( f(x) \). If you're given a function \( h(x) = \frac{g(x)}{f(x)} \), how do you find \( h'(x) \)?

• Start by identifying which functions are your numerator \( g(x) \) and denominator \( f(x) \).


• The quotient rule formula tells us: \[ \left( \frac{g(x)}{f(x)} \right)' = \frac{f(x)g'(x) - g(x)f'(x)}{(f(x))^2}. \]


• This formula helps you to calculate \( h'(x) \) by taking the derivative of \( g(x) \) times the original \( f(x) \), minus \( g(x) \) times the derivative of \( f(x) \). Then divide this result by \( (f(x))^2 \), which is the square of the denominator.

Applying this rule to exercises involving fractions becomes manageable with practice, and it's especially important when functions are used in real-world applications.

Differentiable Functions: The Key to Smooth Calculations

Differentiable functions are functions that have a derivative at every point in their domain. This means the function is smooth and has no sharp edges or discontinuities.
Here's why differentiable functions are crucial:

• They ensure the quotient rule and other differentiation techniques can be applied effectively.


• They guarantee that for any point \( x \), the function has a well-defined derivative \( f'(x) \).
• Understanding how to identify and work with these functions allows smooth and precise calculation of derivatives, particularly when dealing with complex compound functions.

A key characteristic of differentiable functions is that their graph has a tangent line at every point, meaning there's a single, clear slope. This smoothness is why derivatives can be calculated accurately.

Undefined Derivative: When Rules Hit A Wall

An undefined derivative occurs when the rules of differentiation result in a problem, such as division by zero. In the example of \( h(x) = \frac{g(x)}{f(x)} \), if at any value \( x \), \( f(x) = 0 \), the denominator in the quotient rule formula becomes zero, making the derivative undefined.
Here's what's happening:

• When substituting values into the derivative expression, if the denominator results in zero, you're dividing by zero, which is undefined in mathematics.


• This situation occurred at \( x = 4 \) in the exercise, where \( f(4) = 0 \), leading to an undefined derivative \( h'(4) \).
• This warns you to always check the denominator before finalizing derivative calculations to preemptively catch these situations.

Understanding when derivatives don't work out provides a deeper insight into function behavior, identifying critical points where normal rules break and highlighting potential areas for further investigation.