Problem 1
Question
Prove the product rule. Hint: Use \(f(x) g(x)-f(c) g(c)=f(x)(g(x)-g(c))+(f(x)-\) \(f(c)) g(c)\).
Step-by-Step Solution
Verified Answer
By evaluating the limits of the expanded hint expression, we derive the product rule: \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\).
1Step 1: Understand Product Rule
The product rule in calculus tells us how to differentiate a function that is the product of two functions. If you have two functions, \( f(x) \) and \( g(x) \), the derivative of their product is given by \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \).
2Step 2: Expand the Hint Expression
We are given the expression \( f(x)g(x) - f(c)g(c) = f(x)(g(x)-g(c)) + (f(x)-f(c))g(c) \). Let's expand the right-hand side to understand it better: \( f(x)(g(x)-g(c)) + (f(x)-f(c))g(c) = f(x)g(x) - f(x)g(c) + f(x)g(c) - f(c)g(c) \).
3Step 3: Simplify the Expanded Expression
After expanding, we have \( f(x)g(x) - f(c)g(c) \), which simplifies to itself, showing that the expansion is indeed correct. This confirms that the expression holds as presented in the hint.
4Step 4: Apply Limit to Define Derivative
To prove the product rule, take the limit as \( x \to c \): \[ \lim_{x \to c} \frac{f(x)g(x) - f(c)g(c)}{x-c} = \lim_{x \to c} \left(\frac{f(x)(g(x)-g(c))}{x-c} + \frac{(f(x)-f(c))g(c)}{x-c} \right) \].
5Step 5: Evaluate Each Limit Separately
Evaluate the limits separately. The first term becomes \( f(x)g'(x) \) by treating \( f(x) \) as a constant and handling \( g(x)-g(c) \) inside the derivative fraction. The second term turns into \( f'(x)g(x) \) by treating \( g(x) \) as a constant and dealing with the difference quotient for \( f(x) \).
6Step 6: Combine Results to Conclude Proof
Combining the results from the limits gives us the final result: \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \), which is the product rule.
Key Concepts
CalculusDifferentiationReal AnalysisLimit of a Function
Calculus
Calculus is a branch of mathematics focused on studying the rate of change and accumulation. It provides the foundation for understanding changes in functions and introduces the concept of derivatives. Calculus comprises two main components:
It is foundational for many scientific fields like physics, engineering, and economics. Through rules like the product rule, calculus enables us to handle products of functions, making complex equations more approachable.
- Differential Calculus: This section deals with the concept of the derivative, which measures how a function changes as its input changes.
- Integral Calculus: This part involves integration, which is the process of finding the whole from its parts or adding up small data points to calculate the total area under a curve.
It is foundational for many scientific fields like physics, engineering, and economics. Through rules like the product rule, calculus enables us to handle products of functions, making complex equations more approachable.
Differentiation
Differentiation is a method in calculus that involves finding the derivative of a function, which represents the rate at which the function value changes with respect to changes in its input.
The derivative provides insights into a function's instantaneous rate of change and can be used to determine critical points of interest, such as maxima or minima.
The differentiation process requires understanding several rules:
The derivative provides insights into a function's instantaneous rate of change and can be used to determine critical points of interest, such as maxima or minima.
The differentiation process requires understanding several rules:
- Power Rule: For any function of the form \(x^n\), the derivative is \(nx^{n-1}\).
- Product Rule: For differentiating the product of two functions \(f(x)\) and \(g(x)\), use \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \).
- Quotient Rule: When dealing with division of functions, it applies as \( \frac{f(x)}{g(x)}' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \).
Real Analysis
Real analysis provides the rigorous mathematical framework that underpins calculus. It is concerned with the properties of real numbers, sequences, series, and functions.
In real analysis, we explore concepts such as limits, continuity, and convergence, each of which is essential to understanding calculus thoroughly.
Some key topics include:
In real analysis, we explore concepts such as limits, continuity, and convergence, each of which is essential to understanding calculus thoroughly.
Some key topics include:
- Limits: Understanding how functions behave as inputs approach a certain point.
- Continuity: A function is continuous if small changes in input result in small changes in output.
- Convergence: Whether a sequence or a series approaches a specific value as more terms are added.
Limit of a Function
The limit of a function is a fundamental concept in both calculus and real analysis. It describes the behavior of a function as the input approaches a specific value.
Knowing how to evaluate limits is crucial for defining derivatives, integrals, and understanding the overall behavior of functions.
There are important points to consider with limits:
Knowing how to evaluate limits is crucial for defining derivatives, integrals, and understanding the overall behavior of functions.
There are important points to consider with limits:
- Approaching a Point: Limits consider the value a function approaches as the input gets close to a point, not necessarily the value at the point itself.
- L'Hôpital's Rule: Useful for finding limits that result in indeterminate forms like \( \frac{0}{0} \).
- Continuity Link: A function is continuous at a point if the limit at that point equals the function's value there.
Other exercises in this chapter
Problem 1
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuously differentiable such that \(f^{\prime}(x)>0\) for all \(x .\) Show that \(f\) is invertible on t
View solution Problem 1
Compute the nth Taylor Polynomial at 0 for the exponential function.
View solution Problem 2
Suppose I, \(J\) are intervals and a monotone onto \(f: I \rightarrow J\) has an inverse \(g: J \rightarrow I\). Suppose you already know that both \(f\) and \(
View solution Problem 2
Suppose p is a polynomial of degree d. Given any \(x_{0} \in \mathbb{R}\), show that the \((d+1)\) th Taylor polynomial for \(p\) at \(x_{0}\) is equal to \(p\)
View solution