Problem 59 - Thomas Calculus

Question

Suppose that functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=2\) and \(x=3\) . $$ \begin{array}{|c|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 2 & {8} & {2} & {1 / 3} & {-3} \\ \hline 3 & {3} & {-4} & {2 \pi} & {5} \\ \hline\end{array} $$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x .\) $$ \begin{array}{ll}{\text { a. } 2 f(x), \quad x=2} & {\text { b. } f(x)+g(x), \quad x=3} \\ {\text { c. } f(x) \cdot g(x), \quad x=3} & {\text { d. } f(x) / g(x), \quad x=2} \\ {\text { e. } f(g(x)), \quad x=2} & {\text { f. } \sqrt{f(x)}, \quad x=2} \\ {\text { g. } 1 / g^{2}(x), \quad x=3} & {\text { h. } \sqrt{f^{2}(x)+g^{2}(x)}, \quad x=2}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\frac{2}{3}\), (b) \(2\pi + 5\), (c) \(15 - 8\pi\), (d) \(\frac{37}{6}\), (e) \(-1\), (f) \(\frac{1}{12}\sqrt{2}\), (g) \(\frac{5}{32}\), (h) \(\frac{-10}{3\sqrt{68}}\).

Step 1: Differentiate 2f(x) and Evaluate at x=2

The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. Thus, the derivative of \(2f(x)\) is \(2f'(x)\). At \(x=2\), \(f'(2) = \frac{1}{3}\). Consequently, \(2f'(2) = 2 \times \frac{1}{3} = \frac{2}{3}\).

Step 2: Differentiate f(x) + g(x) and Evaluate at x=3

The derivative of the sum of two functions is the sum of the derivatives of the functions. Therefore, \((f(x) + g(x))' = f'(x) + g'(x)\). At \(x=3\), \(f'(3) = 2\pi\) and \(g'(3) = 5\). Thus, \(f'(3) + g'(3) = 2\pi + 5\).

Step 3: Differentiate f(x) * g(x) and Evaluate at x=3

Use the product rule: \((fg)' = f'g + fg'\). At \(x=3\), \(f(3) = 3\), \(g(3) = -4\), \(f'(3) = 2\pi\), and \(g'(3) = 5\). Therefore, \((fg)'(3) = 3 \times 5 + 2\pi \times (-4) = 15 - 8\pi\).

Step 4: Differentiate f(x) / g(x) and Evaluate at x=2

Use the quotient rule: \((\frac{f}{g})' = \frac{f'g - fg'}{g^2}\). At \(x=2\), \(f(2) = 8\), \(g(2) = 2\), \(f'(2) = \frac{1}{3}\), and \(g'(2) = -3\). Therefore, \((\frac{f}{g})'(2) = \frac{(\frac{1}{3} \cdot 2) - (8 \cdot (-3))}{2^2} = \frac{\frac{2}{3} + 24}{4} = \frac{\frac{74}{3}}{4} = \frac{74}{12} = \frac{37}{6}\).

Step 5: Differentiate f(g(x)) and Evaluate at x=2

Use the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). At \(x=2\), \(g(2) = 2\), \(f'(2) = \frac{1}{3}\), and \(g'(2) = -3\). Therefore, \(f'(g(2)) \cdot g'(2) = f'(2) \times -3 = \frac{1}{3} \times -3 = -1\).

Step 6: Differentiate \(\sqrt{f(x)}\) and Evaluate at x=2

The derivative of \(\sqrt{f(x)}\) is \(\frac{1}{2\sqrt{f(x)}} \cdot f'(x)\). At \(x=2\), \(f(2) = 8\) and \(f'(2) = \frac{1}{3}\). Thus, \(\frac{1}{2\sqrt{8}} \cdot \frac{1}{3} = \frac{1}{6\sqrt{8}} = \frac{1}{12}\sqrt{2}\).

Step 7: Differentiate \(\frac{1}{g^2(x)}\) and Evaluate at x=3

Use the chain rule: if \(u = g(x)\), then \(\frac{d}{dx}[\frac{1}{u^2}] = \frac{-2g'(x)}{g^3(x)}\). At \(x=3\), \(g(3) = -4\) and \(g'(3) = 5\). Thus, \(\frac{-2 \cdot 5}{(-4)^3} = \frac{-10}{-64} = \frac{5}{32}\).

Step 8: Differentiate \(\sqrt{f^2(x) + g^2(x)}\) and Evaluate at x=2

Use the chain rule: \(\frac{1}{2\sqrt{f^2(x) + g^2(x)}} \cdot (2f(x)f'(x) + 2g(x)g'(x))\). At \(x=2\), \(f(2) = 8\), \(g(2) = 2\), \(f'(2) = \frac{1}{3}\), \(g'(2) = -3\). Therefore, \(\frac{1}{2\sqrt{64 + 4}} \cdot (\frac{16}{3} - 12) = \frac{1}{2\sqrt{68}} \cdot \left(-\frac{20}{3}\right) = \frac{-10}{3\sqrt{68}}\).

Key Concepts

Product Rule

The product rule is a handy tool for finding the derivative of a product of two functions. When you have a function that is the multiplication of two separate functions, like \( f(x) \) and \( g(x) \), the product rule tells us how to differentiate it. The rule says:

• The derivative of \( f(x) \cdot g(x) \) is given by \((fg)' = f'g + fg'\).
• In simple terms, you take the derivative of the first function, multiply it by the second function, then add it to the first function multiplied by the derivative of the second function.

Let's see how it works with an example: At \( x=3 \), we have \( f(3) = 3 \), \( g(3) = -4 \), \( f'(3) = 2\pi \), and \( g'(3) = 5 \). Then,\[ (fg)'(3) = f'(3) \cdot g(3) + f(3) \cdot g'(3) = 2\pi \cdot (-4) + 3 \cdot 5 = -8\pi + 15 \]The product rule makes calculating these derivatives straightforward.

Chain Rule

The chain rule is a method to find the derivative of compositions of functions. This is useful when one function is inside another function, denoted as \( f(g(x)) \). The chain rule states:

• For the derivative of \( f(g(x)) \), compute \( f'(g(x)) \) multiplied by \( g'(x) \).
• This helps to "chain" the differentiation from the outer function to the inner function.

Consider how this applies to our exercise: At \( x=2 \), for \( f(g(x)) \), where \( g(2) = 2 \), we know \( f'(2) = \frac{1}{3} \) and \( g'(2) = -3 \). Thus, \[ f'(g(2)) \cdot g'(2) = \frac{1}{3} \times (-3) = -1 \]We used the chain rule to seamlessly differentiate the composition, capturing the effect of both functions in the derivative.

Quotient Rule

The quotient rule is designed for finding the derivative of a ratio of two functions. When you have a function in the format of \( \frac{f(x)}{g(x)} \), the quotient rule comes into play. The rule says:

• The derivative of \( \frac{f(x)}{g(x)} \) is \((\frac{f}{g})' = \frac{f'g - fg'}{g^2}\).
• This involves differentiating both the numerator and the denominator, following a specific order.

Applying this at \( x=2 \), where \( f(2) = 8 \), \( g(2) = 2 \), \( f'(2) = \frac{1}{3} \), and \( g'(2) = -3 \):\[ \left(\frac{f}{g}\right)'(2) = \frac{(\frac{1}{3} \cdot 2) - (8 \cdot (-3))}{2^2} = \frac{\frac{2}{3} + 24}{4} = \frac{74}{12} = \frac{37}{6} \]This method ensures we accurately capture how the quotient changes with respect to \( x \).

Sum of Derivatives

This concept deals with how derivatives behave under addition or subtraction. Luckily, this rule is very straightforward, as:

• The derivative of a sum \( f(x) + g(x) \) is the sum of their derivatives: \( (f + g)' = f' + g' \).
• It means you can differentiate each term separately and then add or subtract as needed.

For instance, let's see how it applies to our problem: At \( x=3 \), we know \( f'(x) = 2\pi \) and \( g'(x) = 5 \). Hence,\[ (f(x) + g(x))' = f'(3) + g'(3) = 2\pi + 5 \]This rule simplifies the process when you're working with sums or differences of functions, facilitating easy differentiation.