Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the indicated derivative in terms of the function \(F(x) .\) Assume that \(F\) is differentiable. $$ \frac{d}{d z}(1+(F(2 z)))^{2} $$

Step-by-Step Solution

Answer

\( 4(1 + F(2z))F'(2z) \)

Step 1: Recognize the Type of Derivative

The problem requires finding the derivative of a composition of functions. The outer function is a power, and the inner function involves a composition with a function of two times the variable. We'll use the chain rule for differentiation.

Step 2: Apply the Chain Rule

The chain rule states \( \frac{d}{d z}[u(z)]^n = n[u(z)]^{n-1} \cdot \frac{du}{dz} \). Let \( u(z) = 1 + F(2z) \). Then, \( \frac{d}{d z}(1 + F(2z))^{2} = 2(1 + F(2z)) \cdot \frac{d}{d z}(1 + F(2z)) \).

Step 3: Compute the Inner Derivative

Now focus on \( \frac{d}{d z}(1 + F(2z)) \). This can be broken down by differentiating each component. The derivative of \( 1 \) is \( 0 \), and \( F(2z) \) requires another application of the chain rule. Thus, \( \frac{d}{d z}[F(2z)] = F'(2z) \cdot \frac{d}{d z}(2z) = 2F'(2z) \).

Step 4: Combine the Results

Substitute \( \frac{d}{d z}(1 + F(2z)) = 2F'(2z) \) back into the derivative from Step 2. So, \( \frac{d}{d z}(1 + F(2z))^{2} = 2(1 + F(2z)) \cdot 2F'(2z) = 4(1 + F(2z)) \cdot F'(2z) \).

Step 5: Write the Final Expression

The final expression for the derivative is a function expressed entirely in terms of \( F \) and its derivative: \( 4(1 + F(2z))F'(2z) \).

Key Concepts

Understanding Derivatives

Derivatives play a crucial role in calculus and are essential for understanding how functions behave. At its core, a derivative measures the rate at which a function's value changes as its input changes. If you imagine a curve of a function on a graph, the derivative at any given point is the slope of the tangent line to the curve at that point. This gives us valuable insight into the function's trend—whether it's increasing, decreasing, or constant at any moment.

In mathematical notation, if we have a function \( y = f(x) \), then the derivative of that function with respect to \( x \) is denoted by \( \frac{dy}{dx} \) or \( f'(x) \). The derivative tells us:

• Instantaneous rate of change of the function at a specific point.
• Behavior of the function around that point—peaks, troughs, or constants.

Remember, a derivative is not just a number; it's a function that tells how fast or slow another function is changing. This concept is fundamental for everything from basic motion problems to complex economic models in calculus.

Exploring Composite Functions

Composite functions occur when one function is applied to the result of another function. This layering can make finding derivatives more complex, but it also lets us describe intricate behaviors with simpler functions stacked together.

Let's say we have two functions, \( f(x) \) and \( g(x) \). A composite function is written as \( f(g(x)) \), meaning you first apply \( g \) to \( x \), and then \( f \) to the result of \( g(x) \).

Understanding composition:

• It enables the creation of complex models from simpler ones.
• Composite functions are key in real-world applications, from physics to finance.

When differentiating composite functions, we must carefully account for each layer of function applied. This leads us to a critical differentiation technique known as the chain rule, which beautifully simplifies this complex task.

Differentiation Techniques: The Chain Rule

Differentiating composite functions becomes manageable with the chain rule. It is a powerful tool that allows us to find the derivative of composite functions step-by-step. The rule essentially tells us how functions nested within each other can be unraveled and differentiated.

The chain rule states: If \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). It means:

• First, find the derivative of the outer function \( f \) with respect to the inside function \( g(x) \).
• Then, multiply this by the derivative of the inside function \( g(x) \) with respect to \( x \).

In the given exercise, the chain rule helps us start by identifying the layers. First, differentiate the outer power function, then proceed to find the inner derivative by applying the chain rule again to a nested function \( 1 + F(2z) \). Finally, combining these results snapshot the careful unwinding of composite layers—a great way to handle complex problems smoothly.