Problem 60 - Thomas Calculus

Question

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1\) . $$ \begin{array}{|c|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 0 & {1} & {1} & {5} & {1 / 3} \\ \hline 1 & {3} & {-4} & {-1 / 3} & {-8 / 3} \\ \hline\end{array} $$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x,\) $$ \begin{array}{ll}{\text { a. } 5 f(x)-g(x), \quad x=1} & {\text { b. } f(x) g^{3}(x), \quad x=0} \\ {\text { c. } \frac{f(x)}{g(x)+1}, \quad x=1} & {\text { d. } f(g(x)), \quad x=0} \\ {\text { e. } g(f(x)), \quad x=0} & {\text { f. }\left(x^{11}+f(x)\right)^{-2}, \quad x=1}\end{array} $$ $$ \begin{array}{l}{\text { g. } f(x+g(x)), \quad x=0}\end{array} $$

Step-by-Step Solution

Answer

a) 1, b) 8, c) 17/3, d) 5/3, e) -40/3, f) -0.133, g) -10/9.

Step 1: Differentiate 5f(x) - g(x) and Evaluate at x=1

To find the derivative of \(5f(x) - g(x)\), apply the linearity property of differentiation: \(\frac{d}{dx}[5f(x) - g(x)] = 5f'(x) - g'(x)\). Substitute the values at \(x=1\): \(5(-1/3) - (-8/3) = -5/3 + 8/3 = 3/3 = 1\).

Step 2: Differentiate f(x)g^3(x) and Evaluate at x=0

Use the product rule \(\frac{d}{dx}[f(x)g^3(x)] = f'(x)g^3(x) + f(x)\cdot 3g^2(x)g'(x)\). Substitute the values at \(x=0\): \(5(1)^3 + 1\cdot3(1)^2(1/3) = 5 + 3 = 8\).

Step 3: Differentiate \(\frac{f(x)}{g(x)+1}\) and Evaluate at x=1

Apply the quotient rule: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)+1}\right) = \frac{(g(x)+1)f'(x) - f(x)g'(x)}{(g(x)+1)^2}\). Substitute the values at \(x=1\): \(((-4 + 1)(-1/3) - 3(-8/3))/(3-4)^2 = (-1)(-1/3 - 8)/1 = 17/3\).

Step 4: Differentiate f(g(x)) using the Chain Rule and Evaluate at x=0

Apply chain rule: \(\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)\). Substitute the values at \(x=0\): \(f'(1)\cdot(1/3) = 5\times1/3 = 5/3\).

Step 5: Differentiate g(f(x)) using the Chain Rule and Evaluate at x=0

Apply chain rule: \(\frac{d}{dx}[g(f(x))] = g'(f(x))f'(x)\). Substitute the values at \(x=0\): \(g'(1)\cdot5 = (-8/3)\times5 = -40/3\).

Step 6: Differentiate \((x^{11}+f(x))^{-2}\) and Evaluate at x=1

Apply chain rule: \(\frac{d}{dx}[(x^{11}+f(x))^{-2}] = -2(x^{11}+f(x))^{-3}(11x^{10}+f'(x))\). Substitute at \(x=1\): \(-2(1+3)^{-3}(11+(-1/3)) = -2(4^3)^{-1}(10.67) = -0.133\).

Step 7: Differentiate f(x+g(x)) using the Chain Rule and Evaluate at x=0

Apply chain rule: \(\frac{d}{dx}[f(x+g(x))] = f'(x+g(x))(1 + g'(x))\). Substitute the values at \(x=0\): \(f'(2)(1 + 1/3) = (-1/3)\times(4/3) = -10/9\).

Key Concepts

Product Rule in Differentiation

In calculus, the product rule is a handy tool when you need to differentiate a function that is the product of two simpler functions. If you have two functions, say \( u(x) \) and \( v(x) \), then their product is \( u(x) \cdot v(x) \).

To differentiate this product, you apply the product rule, which states:

• \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

In short, you take the derivative of the first function, multiply it by the second function, then add the first function multiplied by the derivative of the second function.

For example, in the original exercise, we applied the product rule to find the derivative of \( f(x)g^3(x) \). Here, the function \( u(x) \) is \( f(x) \) and \( v(x) \) is \( g^3(x) \), which is a composition of \( g(x) \). This demonstrates how the product rule comes in handy when handling derivatives in calculus.

Quotient Rule Application

The quotient rule is used when dealing with the differentiation of a fraction of two functions. When you have a function \( \frac{u(x)}{v(x)} \), the quotient rule helps you find its derivative efficiently.

The rule states:

• \[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

Essentially, you take the derivative of the numerator, multiply it by the denominator, subtract the product of the numerator and the derivative of the denominator, and finally divide everything by the square of the denominator.

In the problem given, this method was applied to solve part (c) of the exercise, \( \frac{f(x)}{g(x)+1} \). We used the quotient rule to simplify finding the derivative without getting caught up in complex algebraic manipulations.

Chain Rule and Its Importance

The chain rule is a fundamental differentiation technique used to find the derivative of composite functions, i.e., where a function is inside another function. Suppose you have a function \( h(x) = f(g(x)) \), here \( g(x) \) is inside \( f \), making it a composite.

The chain rule states:

• \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

This tells us to find the derivative of the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function.

This concept shines in the solutions for parts (d), (e), and (g) of the exercise where functions like \( f(g(x)) \) and \( g(f(x)) \) were differentiated. Understanding the chain rule is essential because it streamlines the process of dealing with nested functions.

Understanding Function Differentiation

Function differentiation is the process of finding the rate at which a function changes at any given point. Differentiation is the core idea of calculus, expressing velocity, acceleration, and growth in various disciplines.

It involves finding the derivative, often represented as \( f'(x) \), of a function \( f(x) \), capturing the slope of the tangent to the function's graph at a given point. By following the rules of differentiation, namely the product, quotient, and chain rules, you can tackle various types of functions effectively.

In the context of the original exercise, differentiating combinations of functions using these rules allowed us to assess the behavior of functions \( f \) and \( g \) at specific points. Mastering these differentiation techniques opens up a world of applications in mathematics and related fields, making it crucial for higher-level understanding.