Problem 59 - Thomas Calculus in SI Units

Question

Assume that \(f\) and \(g\) are differentiable functions that are inverses of one another so that \((g \circ f)(x)=x .\) Differentiate both sides of this equation with respect to \(x\) using the Chain Rule to express \((g \circ f)^{\prime}(x)\) as a product of derivatives of \(g\) and \(f .\) What do you find? (This is not a proof of Theorem 1 because we assume here the theorem's conclusion that \(g=f^{-1}\) is differentiable.)

Step-by-Step Solution

Answer

\(g'(f(x)) = \frac{1}{f'(x)}\).

Step 1: Understand the Problem

We are given that \((g \circ f)(x) = x\), which means that \(g\) is the inverse of \(f\). Our task is to differentiate \((g \circ f)(x)\) using the Chain Rule and express it in terms of the derivatives of \(g\) and \(f\).

Step 2: Apply the Chain Rule

The Chain Rule states that if \(h(x) = g(f(x))\), then \(h'(x) = g'(f(x)) \cdot f'(x)\). Here, \((g \circ f)(x) = x\), so differentiating both sides with respect to \(x\) gives us \((g \circ f)'(x) = 1\).

Step 3: Differentiate the Composite Function

Using the Chain Rule on \((g \circ f)(x) = x\), we have:\[(g \circ f)'(x) = g'(f(x)) \cdot f'(x)\]According to Step 2, this derivative must equal 1, so:\[g'(f(x)) \cdot f'(x) = 1\]

Step 4: Solve for \(g'(f(x))\)

From the equation \(g'(f(x)) \cdot f'(x) = 1\), solve for \(g'(f(x))\):\[g'(f(x)) = \frac{1}{f'(x)}\]This means that the derivative of the inverse function at \(f(x)\) is the reciprocal of the derivative of the original function at \(x\).

Key Concepts

Inverse Functions

Inverse functions are fascinating! They are like two puzzle pieces that fit together perfectly. When you have a function \(f(x)\) and its inverse \(g(x) = f^{-1}(x)\), applying one followed by the other brings you back to where you started. Simply put, \(f(g(x)) = x\) and \(g(f(x)) = x\). To visualize this, think of them as undoing each other’s actions. If \(f(x)\) shifts \(x\) up by 2, then \(f^{-1}(x)\) will shift it back down by 2. They complement each other perfectly!In calculus, inverse functions help us understand how two variables relate to each other. They are essential, as finding inverses, especially when dealing with trigonometric functions and their inverses, is crucial. Understanding the inverse is key in areas such as solving equations or integrating functions that involve either \(f(x)\) or its inverse \(f^{-1}(x)\).

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative tells us how fast a function is changing at any point—a critical concept in calculus and real-world problem solving. For any function \(f(x)\), its derivative \(f'(x)\) can provide us with the slope of the tangent line at any given point on the function's curve. Differentiation is used to analyze ways in which a function behaves. It helps in determining maxima or minima of a function, which describes critical points where the function’s rate of increase or decrease changes. This plays a role in optimization problems where the goal is to find the maximum or minimum of a function for given constraints.The basic rules of differentiation, like the product rule and the quotient rule, help us to differentiate complex functions step by step. In our problem, the Chain Rule is a vital tool since it allows us to differentiate expressions where one function is inside another, known as a composite function.

Composite Function Differentiation

When we talk about differentiating composite functions, we are dealing with scenarios where one function is plugged into another—just like nested boxes. Using the Chain Rule is crucial here. If you have a function \(h(x) = g(f(x))\), the Chain Rule helps us find the derivative of \(h\) by taking the derivative of \(g\) and \(f\) and multiplying them. The Chain Rule formula is: \[h'(x) = g'(f(x)) \cdot f'(x)\].This notion becomes handy when differentiating inverse functions. Since \((g \circ f)(x) = x\), using the Chain Rule gives us \[g'(f(x)) \cdot f'(x) = 1\]. This equation elegantly states that the derivative of the inverse function at \(f(x)\) is the reciprocal of the derivative of the original function at \(x\). This relationship is pivotal in understanding how inverse functions synchronize their rates of change with their original counterparts.Being comfortable with composite function differentiation opens the door to analyzing more sophisticated problems in calculus, making it a practical skill for any math student to master.