Problem 54 - Thomas Calculus

Question

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=1-\frac{1}{u}, \quad u=g(x)=\frac{1}{1-x}, \quad x=-1 $$

Step-by-Step Solution

Answer

The value of \((f \circ g)'\) at \(x = -1\) is 1.

Step 1: Find the Expression for the Composite Function

First, we need to establish the composition of the functions \( f \) and \( g \). Given \( g(x) = \frac{1}{1-x} \), substitute \( u \) in \( f(u) \) to get the composite function:\[ (f \circ g)(x) = 1 - \frac{1}{g(x)} = 1 - \frac{1}{(\frac{1}{1-x})}. \\]This simplifies to:\[(f \circ g)(x) = 1 - (1-x) = x. \]

Step 2: Differentiate the Composite Function

Now we differentiate the expression \((f \circ g)(x) = x\) with respect to \(x\). Since it simplifies to just \(x\), we have:\[(f \circ g)'(x) = \frac{d}{dx}(x) = 1.\]

Step 3: Evaluate the Derivative at the Given Value of \(x\)

Now we need to evaluate \((f \circ g)^{\prime}(x)\) at the given \(x = -1\):Since the derivative \((f \circ g)^{\prime}(x) = 1\), it is the same for any value of \(x\). Thus, \\[(f \circ g)^{\prime}(-1) = 1.\]

Key Concepts

Function Composition

Function composition is a mathematical process where you combine two functions to create a new one. This involves plugging one function into another. For example, if you have two functions, \( f(u) \) and \( g(x) \), composing these would mean that the output of \( g(x) \) becomes the input for \( f(u) \). So you get a new function: \( (f \circ g)(x) = f(g(x)) \).
In our exercise, we have \( f(u) = 1 - \frac{1}{u} \) and \( u = g(x) = \frac{1}{1-x} \). By substituting \( g(x) \) into \( f(u) \), the composite function becomes \( (f \circ g)(x) = 1 - \frac{1}{g(x)} \).
This simplifies to \( (f \circ g)(x) = x \), showing how g's structure cancels out in the composition.

Derivatives

Derivatives are a fundamental concept in calculus which measure how a function changes as its input changes; essentially, it represents a rate of change. In more familiar terms, derivatives can be thought of as slopes of tangent lines to the function graph at any given point.
To compute a derivative, you apply rules of differentiation to the function's expression. In the exercise, we found the derivative of \( (f \circ g)(x) = x \). Since the expression simplifies directly to \( x \), its derivative is straightforward: \( \frac{d}{dx}(x) = 1 \).
This process shows that no matter the value of \( x \), the rate of change here is constant, reinforcing the simplicity of the linear function in this context.

Chain Rule

The chain rule is a powerful differentiation rule for computing the derivative of a composite function. When you have a function composed of two or more functions, the chain rule provides a systematic method to differentiate it. Essentially, it states that the derivative of \( f(g(x)) \) is the derivative of \( f \) with respect to \( g \) times the derivative of \( g \) with respect to \( x \). Mathematically, this is expressed as \( (f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x) \).
In our exercise, although it appears straightforward, the chain rule underpins the simplification. The construction of the composite function eliminates complexity before differentiation due to the specific behavior of the given functions ensuring even without complex applications, the basis remains the systematic process embedded within the chain rule.
Understanding this equips you to handle more intricate compositions where direct simplification might not always occur effortlessly.