Problem 129 - Calculus Volume 1

Question

For the following exercises, 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. $$\begin{array}{|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} \\ \hline f(x) & {3} & {5} & {-2} & {0} \\ \hline g(x) & {2} & {3} & {-4} & {6} \\ \hline f^{\prime}(x) & {-1} & {7} & {8} & {-3} \\ \hline g^{\prime}(x) & {4} & {1} & {2} & {9} \\ \hline\end{array}$$ 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 because \(f(4) = 0\).

Step 1: Differentiate the Expression

To find \(h'(x)\), we start by differentiating \(h(x) = \frac{1}{x} + \frac{g(x)}{f(x)}\). The derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\). For \(\frac{g(x)}{f(x)}\), we apply the quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\), where \(u(x) = g(x)\) and \(v(x) = f(x)\).

Step 2: Apply the Quotient Rule

For the term \(\frac{g(x)}{f(x)}\), differentiate using the quotient rule. Let \(u(x) = g(x)\) and \(v(x) = f(x)\). Therefore, \(u' = g'(x)\) and \(v' = f'(x)\). The derivative is \(h'(x) = -\frac{1}{x^2} + \frac{g'(x)f(x) - g(x)f'(x)}{(f(x))^2}\).

Step 3: Substitute Values for x=4

Using the table, substitute values at \(x=4\): \(f(4)=0\), \(g(4)=6\), \(f'(4)=-3\), \(g'(4)=9\). Substituting these into the derivative, we get \(h'(4) = -\frac{1}{4^2} + \frac{9 \cdot 0 - 6 \cdot (-3)}{0^2}\).

Step 4: Evaluate the Expression

Evaluate \(h'(4) = -\frac{1}{16} + \frac{0 + 18}{0}\). The function \(h(x)\) includes division by \(f(x)\) in \(\frac{g(x)}{f(x)}\); since \(f(4) = 0\), this causes the derivative to be undefined at \(x=4\).

Key Concepts

Differentiable functions

In calculus, a function is said to be differentiable at a point if it has a derivative at that point. This means the function can be smoothly traced without any sharp corners or breaks at the point in question. Differentiability implies continuity, but not all continuous functions are differentiable. For example, functions like the absolute value of x are continuous everywhere but not differentiable at a sharp turn, like at zero.
- In this exercise, both functions, denoted as \(f(x)\) and \(g(x)\), are differentiable. This means we can find their derivatives, denoted as \(f'(x)\) and \(g'(x)\), respectively, which are provided in the given table.
- The derivative provides information about the rate at which a function's value is changing at any given point, resembling the slope of the tangent line at that point.
- Since we have the derivatives \(f'(x)\) and \(g'(x)\) available at specified values in the table, we can utilize this information to compute the derivative of a composed function involving \(f(x)\) and \(g(x)\), like \(h(x)\) in our problem.

Derivative calculation

Derivative calculation is a fundamental concept in calculus that allows us to determine how a function changes at any point. This involves finding the derivative of the function, which can provide us with the slope or rate of change of the function.
- For simple functions like polynomials and powers of x, derivatives are calculated using power rules where the exponent of x becomes a coefficient and is reduced by 1.
- For more complex expressions, we might use rules like the product rule, the quotient rule, or the chain rule. Each of these rules allows us to systematically find derivatives for combined functions based on the derivatives of their simpler components.
- In the exercise, we calculate the derivative of \(h(x) = \frac{1}{x} + \frac{g(x)}{f(x)}\). This involves techniques like directly differentiating \(\frac{1}{x}\) and utilizing the quotient rule for the second term involving a ratio of differentiable functions \(\frac{g(x)}{f(x)}\).

Quotient rule

The quotient rule is a technique for finding the derivative of a ratio of two differentiable functions. It's particularly useful when dealing with fractions that involve both the numerator and the denominator as functions of x.
- Mathematically, for two functions \(u(x)\) and \(v(x)\), the derivative of their quotient, \(\frac{u}{v}\), is given by \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). This formula requires both \(u\) and \(v\) to be differentiable, and \(v(x)eq0\) to avoid division by zero.
- Applying this rule to the function \(h(x) = \frac{1}{x} + \frac{g(x)}{f(x)}\), we set \(u(x) = g(x)\) and \(v(x) = f(x)\) for the second term. Then, \(u' = g'(x)\) and \(v' = f'(x)\).
- The derivative of this term using the quotient rule is \(\frac{g'(x)f(x) - g(x)f'(x)}{(f(x))^2}\), which is a key step in determining \(h'(x)\).

Undefined derivative

An undefined derivative occurs when calculating a derivative involves division by zero. This can happen in scenarios involving the quotient rule if the denominator, after substitution of specific values, results in zero.
- In our exercise, we substitute values into \(h'(x)\) to find \(h'(4)\). Using the table, at \(x=4\), \(f(4) = 0\) is substituted as the denominator in the quotient rule term, leading to division by zero and rendering the derivative undefined.
- Undefined derivatives often indicate points of non-differentiability in the original function at those specific x-values. For \(h(x)\), because \(\frac{g(x)}{f(x)}\) is undefined at \(x=4\) due to \(f(4) = 0\), we encounter such a situation.
- It's important to consider these cases, as they highlight moments where the smooth behavior of a function might be disrupted.