Problem 84 - Brief Applied Calculus

Question

The Chain Rule (page 155 ) states that the derivative of \(f(g(x))\) is \(f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Use Carathéodory's definition of the derivative to prove the Chain Rule by giving reasons for the following steps. a. Since \(g\) is differentiable at \(x\), there is a function \(G\) that is continuous at 0 and such that \(g(x+h)-g(x)=G(h) \cdot h\), and \(G(0)=g^{\prime}(x)\). b. Since \(f\) is differentiable at \(g(x)\), there is a function \(F\) that is continuous at 0 and such that \(f(g(x)+h)-f(g(x))=F(h) \cdot h\), and \(F(0)=f^{\prime}(g(x))\) c. For the function \(f(g(x))\) we have \(f(g(x+h))-f(g(x))\) \(\quad=f(g(x)+g(x+h)-g(x))-f(g(x))\) \(\quad=f(g(x)+(g(x+h)-g(x)))-f(g(x))\) \(=F(g(x+h)-g(x)) \cdot(g(x+h)-g(x))\) \(=F(g(x+h)-g(x)) \cdot G(h) \cdot h\) Therefore, the derivative of \(f(g(x))\) is \(\begin{aligned} F(g(x+0)-g(x)) \cdot G(0) &=F(0) \cdot G(0) \\ &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \end{aligned}\) as was to be proved.

Step-by-Step Solution

Answer

The derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\).

Step 1: Understanding Differentiability of g

Since the function \(g\) is differentiable at \(x\), we can express \(g(x+h)\) in terms of a continuous function \(G\). Specifically, \(g(x+h) - g(x) = G(h) \cdot h\), where \(G(0) = g'(x)\). This means that the change in \(g\) with respect to \(h\) can be modeled as a product of a small change factor \(G(h)\) and the increment \(h\), showing how \(g\) behaves approximately linearly at \(x\).

Step 2: Understanding Differentiability of f

Similarly, because \(f\) is differentiable at \(g(x)\), we introduce a function \(F\) such that the change in \(f\) can be represented as \(f(g(x)+h) - f(g(x)) = F(h) \cdot h\), with \(F(0) = f'(g(x))\). This shows that the behavior of \(f\) near \(g(x)\) can be linearly approximated with respect to \(h\).

Step 3: Substitution into the Composite Function

For the composite function \(f(g(x))\), the change \(f(g(x+h)) - f(g(x))\) can be expressed using the substitutions from steps 1 and 2. We substitute \(g(x+h) - g(x)\) from Step 1 into the expression for \(f\). This gives us: \(f(g(x) + (g(x+h) - g(x))) - f(g(x)) = F(g(x+h) - g(x)) \cdot (g(x+h) - g(x))\).

Step 4: Applying the Functions F and G

Using the substitution from Step 1, we substitute \(g(x+h) - g(x) = G(h) \cdot h\) into the expression derived in Step 3. Therefore, \(f(g(x+h)) - f(g(x)) = F(g(x+h) - g(x)) \cdot G(h) \cdot h\).

Step 5: Finding the Derivative using Continuity at Zero

To find the derivative at \(h = 0\), observe that \(F(g(x+0) - g(x)) = F(0)\) and \(G(0) = g'(x)\). Hence, \(f(g(x+h)) - f(g(x)) \approx F(0) \cdot G(0) \cdot h\), showing that \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).

Key Concepts

Derivatives

Let's dive into what derivatives really are. Derivatives help us understand how a function changes at any given point. To put it simply, a derivative gives us the slope of the tangent line to the graph of the function at a specific point. It tells us the rate at which the function's value is changing with respect to changes in the input variable.
When dealing with derivatives, we often use the notation \( f'(x) \) to represent the derivative of the function \( f(x) \). This notation tells us how \( f(x) \) changes as \( x \) changes.

• Derivatives are crucial in calculus and have applications in physics, engineering, economics, and many other fields.
• To calculate a derivative, you typically need to apply specific rules, such as the power rule, product rule, quotient rule, or chain rule.

The chain rule, for instance, is especially useful when dealing with composite functions, where one function is within another. Its primary role is to help differentiate functions like \( f(g(x)) \). This is done by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

Composite Functions

Composite functions integrate one function into another, forming an amalgam that looks like \( f(g(x)) \). Imagine composite functions as a scenario where the output of one function becomes the input for another.
This forms a chain of operations—hence the need for the chain rule to differentiate them. When we deal with composite functions, identifying each part separately is vital:

• Identify the outer function: This is usually represented as \( f \) in \( f(g(x)) \).
• Find the inner function: This is represented as \( g \), and it's the core of our focus for initial differentiation.

When differentiating composite functions, consider that you're peeling back layers: differentiate the outer function while imagining the inner function as one variable.
Then, differentiate the inner function itself. Fit these together, just like the pieces of a puzzle, to apply the derivatives together and obtain the comprehensive derivative of the composite function.

Differentiability

Differentiability is a property that tells us whether a function \( f(x) \) has a derivative at a particular point. It's like checking whether a function is smooth or whether you can draw a tangent at any given point in a straight, unbroken manner.

• If a function is differentiable at a point, it means you can compute its derivative there, indicating a well-defined instantaneous rate of change.
• However, if a function has sharp corners, cusps, or vertical tangents at certain points, it might not be differentiable there.

Differentiability implies continuity, but not the other way around. A function must be continuous at a point to be differentiable there, but just being continuous doesn't ensure differentiability.
Differentiability becomes complex when dealing with composite functions. Here, both the inner function and the outer function must be differentiable. Therefore, understanding differentiability ensures the application of the chain rule seamlessly, confirming all pieces fit together smoothly like a well-oiled machine.