Problem 103
Question
Tangents Suppose that \(u=g(x)\) is differentiable at \(x=1\) and that \(y=f(u)\) is differentiable at \(u=g(1)\) . If the graph of \(y=f(g(x))\) has a horizontal tangent at \(x=1,\) can we conclude anything about the tangent to the graph of \(g\) at \(x=1\) or the tangent to the graph of \(f\) at \(u=g(1) ?\) Give reasons for your answer.
Step-by-Step Solution
Verified Answer
We cannot conclusively determine which tangent is horizontal; either \( g'(1) = 0 \), \( f'(g(1)) = 0 \), or both.
1Step 1: Define the Concepts
We know that if the graph of a function has a horizontal tangent at a point, the derivative of that function at that point is zero. Here, the function is composed of two functions, where the outer function is differentiated with respect to the inner function, and the inner function is differentiated with respect to its variable.
2Step 2: Apply the Chain Rule
For the function composition \( y = f(g(x)) \), the derivative using the chain rule is \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx} \). We are told \( \frac{dy}{dx} \) at \( x = 1 \) is zero because it has a horizontal tangent there.
3Step 3: Analyze the Condition
Given \( \frac{dy}{dx} = 0 \), applying the chain rule gives us \( \frac{df}{du} \cdot \frac{dg}{dx} = 0 \) at \( x = 1 \). For this product to be zero, either \( \frac{df}{du} = 0 \) or \( \frac{dg}{dx} = 0 \), or both can be zero. Therefore, we cannot conclude which specific derivative is zero just from given information.
4Step 4: Interpret the Implication
Since \( \frac{dy}{dx} = 0 \) results from the product \( \frac{df}{du} \cdot \frac{dg}{dx} \), without additional information about \( \frac{df}{du} \) or \( \frac{dg}{dx} \), it indicates that at least one of these derivatives must be zero. However, it does not specify which one is zero specifically or whether both are.
Key Concepts
Chain RuleDifferentiabilityComposite Functions
Chain Rule
The chain rule is a fundamental concept in calculus crucial for finding the derivative of composite functions. Imagine you're peeling layers of an onion. Each layer represents a function within another function. To differentiate effectively, you unpack one layer at a time.
For a composite function like \( y = f(g(x)) \), the chain rule tells us how to find \( \frac{dy}{dx} \). It states that the derivative of \( y \) with respect to \( x \) is the derivative of \( f \) with respect to \( u \) (denoted \( \frac{df}{du} \)), multiplied by the derivative of \( g \) with respect to \( x \) (denoted \( \frac{dg}{dx} \)). So, the chain rule formula is:
\[ \frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx} \]
This principle allows us to differentiate even the most complex functions by breaking them down into simpler parts. Remember, it's like climbing a ladder, where each step depends on the last.
For a composite function like \( y = f(g(x)) \), the chain rule tells us how to find \( \frac{dy}{dx} \). It states that the derivative of \( y \) with respect to \( x \) is the derivative of \( f \) with respect to \( u \) (denoted \( \frac{df}{du} \)), multiplied by the derivative of \( g \) with respect to \( x \) (denoted \( \frac{dg}{dx} \)). So, the chain rule formula is:
\[ \frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx} \]
This principle allows us to differentiate even the most complex functions by breaking them down into simpler parts. Remember, it's like climbing a ladder, where each step depends on the last.
Differentiability
Differentiability at a point means a function has a derivative at that point. Simply put, the function must be smooth enough without any breaking or sharp turns. Think of it as a regular, well-paved road where you can easily calculate the slope at any given point.
For tackle this, consider:
It is important to realize that differentiability implies continuity. That means for a function to be differentiable, it cannot have holes or jumps in its graph where you’re differentiating.
For tackle this, consider:
- If \( g(x) \) is differentiable at \( x = 1 \), then \( \frac{dg}{dx} \) exists at that point.
- Similarly, if \( f(u) \) is differentiable at \( u = g(1) \), then \( \frac{df}{du} \) exists at that point.
It is important to realize that differentiability implies continuity. That means for a function to be differentiable, it cannot have holes or jumps in its graph where you’re differentiating.
Composite Functions
Composite functions involve one function inside another. It's like a nested doll where each doll fits perfectly inside the larger one. By understanding composite functions, you can simplify complex problem-solving.
The function \( y = f(g(x)) \) is a great example, with \( g \) fitting inside \( f \).
These functions often crop up in real-world models and complex calculus problems. Understanding them helps you break down and analyze larger equations one step at a time, making each part manageable and less intimidating.
The function \( y = f(g(x)) \) is a great example, with \( g \) fitting inside \( f \).
- \( g(x) \) plays the role of the inner function.
- \( f(u) \) acts as the outer function, where \( u = g(x) \).
These functions often crop up in real-world models and complex calculus problems. Understanding them helps you break down and analyze larger equations one step at a time, making each part manageable and less intimidating.
Other exercises in this chapter
Problem 101
Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and t
View solution Problem 102
Chain Rule Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the composites $$ \begin{aligned}(f \circ g)(x)=|x|^{2}=x^{2} & \text { and } \quad(g \circ f)(x)
View solution Problem 104
Suppose that \(u=g(x)\) is differentiable at \(x=-5, y=f(u)\) is differentiable at \(u=g(-5),\) and \((f \circ g)^{\prime}(-5)\) is negative. What, if anything,
View solution Problem 105
The derivative of \(\sin 2 x\) Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$ y=\frac{\sin 2(x+h)-\sin 2 x}
View solution