Question

Derivatives from a table \(U\) se the following table to find the given derivatives. $$\begin{array}{lllll} x & 1 & 2 & 3 & 4 \\ \hline f(x) & 5 & 4 & 3 & 2 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 \\ g(x) & 4 & 2 & 5 & 3 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 \end{array}$$ $$\left.\frac{d}{d x}\left(\frac{x f(x)}{g(x)}\right)\right|_{x=4}$$

Short answer

Question: Determine the value of the derivative \(\frac{d}{d x}\left(\frac{x f(x)}{g(x)}\right)\) at x=4, given the table of functions \(f(x)\), \(f'(x)\), \(g(x)\), and \(g'(x)\). Answer: The value of the derivative \(\frac{d}{d x}\left(\frac{x f(x)}{g(x)}\right)|_{x=4} = \frac{14}{9}\).

Step-by-step solution

Applying the Quotient Rule

To find the derivative of the given function \(\frac{x f(x)}{g(x)}\), we will use the quotient rule, which states that if \(u(x) = \frac{v(x)}{w(x)}\), then \(u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2}\). Here, \(v(x) = x f(x)\) and \(w(x) = g(x)\). So, our task is to find \(v'(x)\) and \(w'(x)\) which we will then use to compute the derivative \(u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2}\).

Finding \(v'(x)\) and \(w'(x)\) using Product Rule and given table details

To find \(v'(x)\), we can use the product rule since \(v(x) = x f(x)\). The product rule states that if \(h(x) = p(x)q(x)\), then \(h'(x) = p'(x)q(x) + p(x)q'(x)\). Applying the product rule, \(v'(x) = (1) f(x) + x f'(x)\) where x = 4. From the given table: \(f(4) = 2\), and \(f'(4) = 1\). So, substituting these values, we get \(v'(4) = (1)(2) + (4)(1) = 6\). Similarly, we are provided with \(g(x)\) and \(g'(x)\), we can directly use the table values to find \(w'(4)\). From the table: \(g(4) = 3\) and \(g'(4) = 1\). Hence, \(w(4) = 3\) and \(w'(4) = 1\).

Applying the Quotient Rule and computing the derivative

Now that we have the values of \(v'(4)\), \(w(4)\), and \(w'(4)\), we can use the quotient rule to find the value of the derivative at x = 4. So, \(u'(4) = \frac{v'(4)w(4) - v(4)w'(4)}{[w(4)]^2} = \frac{(6)(3) - (4)(1)}{(3)^2} = \frac{18 - 4}{(9)} = \frac{14}{9}\). Therefore, the value of the derivative \(\frac{d}{d x}\left(\frac{x f(x)}{g(x)}\right)|_{x=4} = \frac{14}{9}\).

Key concepts

Derivative of a Quotient

When faced with the task of differentiating a function that is the quotient of two other functions, we apply the quotient rule. This rule is a critical concept in calculus, especially when dealing with complex functions not easily separable into simpler parts.

Here's the rule stated simply: if you have a function \(u(x) = \frac{v(x)}{w(x)}\), where both \(v\) and \(w\) are differentiable functions of \(x\), then the derivative of \(u\), denoted \(u'(x)\), is given by \(u'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2}\).

Understanding the quotient rule is essential as it provides a systematic method for finding derivatives of ratios, ensuring accuracy and saving time as opposed to attempting to manually manipulate the function into a more 'derivative-friendly' form.

Product Rule

Just as divisions have their specific rule for differentiation, so do multiplications. The product rule is used when you need to find the derivative of a function that is a product of two other functions. This rule states that if you have \(h(x) = p(x)q(x)\), then the derivative \(h'(x)\) is given by \(h'(x) = p'(x)q(x) + p(x)q'(x)\).

When applied, this rule will require us to differentiate each function separately (first \(p\) and then \(q\)) and then combine the results as outlined. Understanding how to effectively apply the product rule is crucial for handling a wide array of problems in calculus, as many functions are presented in a multiplicative format.

Differentiation from a Table

On many occasions, specifically in applied mathematics or when working with experimental data, we are provided with tabulated values of functions and their derivatives at certain points. Differentiation from a table involves taking these given values and using them directly or in conjunction with differentiation rules to find derivatives at particular points.

This process demonstrates the practical application of calculus concepts where you may not have the functional form of an equation, but you still can analyze and derive rates of change. It can prove vital, for instance, in predicting and understanding real-world phenomena where obtaining an exact mathematical function is difficult or impossible.

Applying the Quotient Rule

To bring the abstract concept of the quotient rule into a practical context, let's consider how to apply it to an actual problem. Applying the quotient rule involves identifying your \(v(x)\) and \(w(x)\) correctly, differentiating both functions (oftentimes using other differentiation rules like the product rule), and then plugging these derivatives into the quotient rule formula.

The process requires careful attention to algebraic detail—such as simplifying expressions and properly handling the subtraction—in order to arrive at the correct derivative. Once mastered, it enables students to tackle a diverse range of problems, reinforcing the understanding of how functions behave when they are interrelated through division.