Problem 59 - Thomas Calculus

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

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

Step 1: Identify the Function Composition

We are given that \((g \circ f)(x) = x\), which means that applying function \(f\) to \(x\) and then applying function \(g\) to the result gives us \(x\). This reflects the property of inverse functions.

Step 2: Differentiate Both Sides with Respect to x

Use the chain rule to differentiate both sides of the equation \((g \circ f)(x) = x\). Differentiating the right side with respect to \(x\) gives 1, since the derivative of \(x\) with respect to \(x\) is 1.

Step 3: Apply the Chain Rule on the Left Side

For the left side, which is \((g \circ f)(x)\), apply the chain rule for differentiation. This gives us \(g'(f(x)) \cdot f'(x)\).

Step 4: Equate the Derivatives

Set the derivative from the chain rule \(g'(f(x)) \cdot f'(x)\) equal to the derivative from the right side, which is 1: \[ g'(f(x)) \cdot f'(x) = 1 \]

Step 5: Express the Relationship Between Derivatives

From the equation \(g'(f(x)) \cdot f'(x) = 1\), solve for one of the derivatives. This gives us: \[ g'(f(x)) = \frac{1}{f'(x)} \] This equation expresses \(g'(f(x))\) as the reciprocal of \(f'(x)\).

Key Concepts

Understanding the Chain Rule

The Chain Rule is an essential concept in calculus that helps in differentiating compositions of functions. To put it simply, it states how to take the derivative of a composite function. A composite function refers to a function created by applying one function to the result of another function. For example, if you have two functions, such as \(f\) and \(g\), the composite function could be represented as \((g \circ f)(x)\).

• When tasked with differentiating \((g \circ f)(x) = x\), we use the chain rule.
• This rule tells us that the derivative \((g \circ f)'(x)\) should be written as \(g'(f(x)) \cdot f'(x)\).
• This means you differentiate the inner function \(f\), and then the outer function \(g\), evaluated at \(f(x)\), and take their product.

Applying the chain rule requires careful attention to both the order in which you apply the functions and their respective derivatives. It helps in breaking down more complicated problems into simpler parts that are easier to solve.

Exploring Derivatives

Derivatives represent the rate of change of a function concerning a variable. In simple terms, they show how a function's value changes as its input changes. For instance, if \(y = f(x)\), then the derivative \(f'(x)\) or \(\frac{dy}{dx}\) tells us the slope of the function \(y\) at any point \(x\).

• The derivative of a constant function, like \(x\) in our case, is always zero unless regarding \(x\) itself, which is 1.
• In this exercise, we take the derivative of \((g \circ f)(x) = x\) and find it to be 1.
• Derivatives allow us to express the relationship between the rates of change of the two functions \(f\) and \(g\).

By differentiating both sides, we conclude with an important result: \(g'(f(x)) \cdot f'(x) = 1\). This signifies that although \(f\) and \(g\) are functions of each other, their respective rates of change are inversely related.

Inverse Function Theorem Insights

The Inverse Function Theorem deals with functions that have inverses, providing conditions under which a given function's inverse is also differentiable. In our context, this theorem confirms that if \(f\) is differentiable and its inverse \(g\) exists, then \(g\) is differentiable as well. Also, it gives us a formula to find the derivative of the inverse function.

• According to this theorem, \(g'(f(x)) = \frac{1}{f'(x)}\).
• This relationship highlights that the derivative of the inverse function is the reciprocal of the original function's derivative.
• It establishes a crucial connection between the two reversely related function derivatives.

The exercise demonstrates this theorem's application without formally proving it, assuming differentiability from the start. The result \(g'(f(x)) = \frac{1}{f'(x)}\) shows that even though we started with an assumption, the mathematics behind the differentiation supports the inverse function theorem's conclusions.