Chapter 3: Problem 51
Express the indicated derivative in terms of the function \(F(x) .\) Assume that \(F\) is differentiable. $$ \frac{d}{d z}(1+(F(2 z)))^{2} $$
Short Answer
Expert verified
\( 4(1 + F(2z))F'(2z) \)
Step by step solution
01
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.
02
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, \( \frac{d}{d z}(1 + F(2z))^{2} = 2(1 + F(2z)) \cdot \frac{d}{d z}(1 + F(2z)) \).
03
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, \( \frac{d}{d z}[F(2z)] = F'(2z) \cdot \frac{d}{d z}(2z) = 2F'(2z) \).
04
Combine the Results
Substitute \( \frac{d}{d z}(1 + F(2z)) = 2F'(2z) \) back into the derivative from Step 2. So, \( \frac{d}{d z}(1 + F(2z))^{2} = 2(1 + F(2z)) \cdot 2F'(2z) = 4(1 + F(2z)) \cdot F'(2z) \).
05
Write the Final Expression
The final expression for the derivative is a function expressed entirely in terms of \( F \) and its derivative: \( 4(1 + F(2z))F'(2z) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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:
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 \( x \).