Problem 30 - The Calculus with Analytic Geometry

Question

Suppose that \(f\) and \(g\) are functions such that \(f^{\prime}(x)=1 / x\) and \(f(g(x))=x\). Prove that if \(g^{\prime}(x)\) exists then \(g^{\prime}(x)=g(x)\).

Step-by-Step Solution

Answer

g'(x) = g(x)

Step 1: Differentiate both sides of the equation

Given that f(g(x)) = x, differentiate both sides with respect to x using the chain rule. This gives us: \[ \frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x] \].

Step 2: Apply the chain rule

Applying the chain rule to the left-hand side, we get: \[f^{\text{'}}(g(x)) \cdot g^{\text{'}}(x) = 1 \].

Step 3: Substitute the given derivative of f

Given that f^{\text{'}}(x) = \frac{1}{x}, substitute g(x) in place of x: \[ \frac{1}{g(x)} \cdot g^{\text{'}}(x) = 1 \].

Step 4: Solve for g'(x)

Solve the equation for g^{\text{'}}(x), giving: \[ g^{\text{'}}(x) = g(x) \].

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function changes as its input changes. It is a fundamental tool used to investigate the behavior of functions and their graphs. When you differentiate a function, you are essentially finding this rate of change.

For instance, if we have a function described as f(x), its derivative, denoted as f′(x), tells us how f(x) changes concerning x.

In the given exercise, we are given that the derivative of the function f, written as f′(x), is equal to 1/x. This piece of information is crucial for us to prove the relationship involving the function g.

Function Composition

Function composition involves combining two functions where the output of one function becomes the input of another. Suppose you have two functions f and g. The composition of these functions is written as f(g(x)). This composition means you first apply g to x and then apply f to the result of g(x).

In the context of our exercise, we are given that f(g(x)) = x. This relationship is integral to proving how derivatives of composed functions relate to each other.
Function composition is a powerful concept in calculus as it allows us to explore more complex scenarios by combining simpler functions.

Calculus Proof

A calculus proof is a logical argument that uses principles of calculus to establish the truth of a mathematical statement. In the given exercise, we need to prove that if g' (x) exists, then it equals g(x). Let's walk through the proof step by step.

First, we start with the given equation f(g(x)) = x. Differentiating both sides with respect to x, we use the chain rule. The differentiation of f(g(x)) with respect to x is f'(g(x))·g'(x), and the differentiation of x with respect to x is 1. Therefore, we get: \[ f^{\text{'}}(g(x)) \times g^{\text{'}}(x) = 1 \]

Next, we substitute the given derivative of f from f'(x) = 1/x. Instead of x, we use g(x) as the argument for f, thus yielding: \[ \frac{1}{g(x)} \times g^{\text{'}}(x) = 1 \]

Finally, solving for g'(x), we multiply both sides of the equation by g(x), resulting in: \[ g^{\text{'}}(x) = g(x) \]

This concludes our proof, and we have successfully demonstrated that g'(x) = g(x) given the initial conditions.