Problem 10
Question
Confirm that the stated formula is the local linear approximation at \(x_{0}=0\) $$ \ln (1+x) \approx x $$
Step-by-Step Solution
Verified Answer
Yes, \( \ln(1+x) \approx x \) is the local linear approximation at \( x_0 = 0 \).
1Step 1: Identify the Function
The function given is \( f(x) = \ln(1+x) \). We want to find its local linear approximation at \( x_0 = 0 \). This means we need to find the tangent line of \( f(x) \) at \( x = 0 \).
2Step 2: Determine the Derivative
Find the first derivative of \( f(x) = \ln(1+x) \). Using the chain rule, the derivative is \( f'(x) = \frac{1}{1+x} \).
3Step 3: Evaluate at \(x_0=0\)
Calculate \( f(0) \) and \( f'(0) \). We have \( f(0) = \ln(1+0) = \ln(1) = 0 \) and \( f'(0) = \frac{1}{1+0} = 1 \).
4Step 4: Use the Linear Approximation Formula
The formula for the local linear approximation of a function \( f(x) \) at \( x_0 \) is given by: \( L(x) = f(x_0) + f'(x_0)(x - x_0) \). Substituting our values, \( L(x) = 0 + 1(x - 0) = x \).
5Step 5: Compare with Given Approximation
The linear approximation \( L(x) = x \) matches the given approximation \( \ln(1+x) \approx x \) at \( x_0 = 0 \). Therefore, the formula is confirmed as the local linear approximation.
Key Concepts
Understanding DerivativesTangent Line and Its ImportanceChain Rule and Its Role in Derivatives
Understanding Derivatives
A derivative represents how a function changes at any given point. Think of it as the function's rate of change. In mathematics, the derivative of a function at a particular point gives us the slope of the tangent line at that point. It's like understanding how steep a hill is at a specific point on the hill.
In the exercise, the function is \( f(x) = \ln(1+x) \), and we need to find its derivative to understand how it behaves around \( x = 0 \). When we calculate the derivative of \( \ln(1+x) \), we obtain \( f'(x) = \frac{1}{1+x} \). This result tells us how the function changes as \( x \) changes, specifically how steep the line is at different points.
In the exercise, the function is \( f(x) = \ln(1+x) \), and we need to find its derivative to understand how it behaves around \( x = 0 \). When we calculate the derivative of \( \ln(1+x) \), we obtain \( f'(x) = \frac{1}{1+x} \). This result tells us how the function changes as \( x \) changes, specifically how steep the line is at different points.
- The derivative \( f'(x) = \frac{1}{1+x} \) shows that as \( x \) increases, the rate of change decreases.
- At \( x = 0 \), \( f'(0) = 1 \), meaning the slope is 1. This slope is used in our local linear approximation, indicating the function increases by 1 unit for every 1 unit increase in \( x \) around \( x = 0 \).
Tangent Line and Its Importance
A tangent line is a straight line that touches a curve at a single point and has the same slope as the curve at that point. In simpler words, it's the best straight-line approximation of the curve near the point of tangency. This concept is crucial for understanding local linear approximation.
In our case, we are focusing on the tangent line to the function \( f(x) = \ln(1+x) \) at the point \( x = 0 \). The slope of this tangent line is the derivative evaluated at \( x = 0 \), which we found to be 1. Thus, the equation of our tangent line, and hence the local linear approximation, is \( L(x) = x \).
In our case, we are focusing on the tangent line to the function \( f(x) = \ln(1+x) \) at the point \( x = 0 \). The slope of this tangent line is the derivative evaluated at \( x = 0 \), which we found to be 1. Thus, the equation of our tangent line, and hence the local linear approximation, is \( L(x) = x \).
- The tangent line simplifies complex functions into linear ones, making it easier to predict and calculate values close to the point of tangency.
- At \( x=0 \), the tangent line perfectly approximates \( \ln(1+x) \) with \( L(x) = x \).
Chain Rule and Its Role in Derivatives
The chain rule is a fundamental technique in calculus used to differentiate composite functions. It allows us to find the derivative of composite functions by connecting the derivatives of the outer and inner functions.
In solving this exercise, we applied the chain rule to the function \( f(x) = \ln(1+x) \). Here, the function inside the logarithm, \( u(x) = 1+x \), is differentiated first, yielding \( u'(x) = 1 \). Then, since the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \), we multiply these derivatives together to get \( \frac{1}{1+x} \).
In solving this exercise, we applied the chain rule to the function \( f(x) = \ln(1+x) \). Here, the function inside the logarithm, \( u(x) = 1+x \), is differentiated first, yielding \( u'(x) = 1 \). Then, since the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \), we multiply these derivatives together to get \( \frac{1}{1+x} \).
- The chain rule helps break down the problem into manageable steps, ensuring precise derivative computation.
- Understanding and applying the chain rule correctly is essential for handling more complex functions and mastering calculus.
Other exercises in this chapter
Problem 9
Find \(d y / d x\) by implicit differentiation. \(\sin \left(x^{2} y^{2}\right)=x\)
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Find \(d y / d x\) $$ y=(\ln x)^{3} $$
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Find the limits. $$ \lim _{t \rightarrow 0} \frac{t e^{t}}{1-e^{t}} $$
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Suppose that \(z=x^{3} y^{2},\) where both \(x\) and \(y\) are changing with time. At a certain instant when \(x=1\) and \(y=2, x\) is decreasing at the rate of
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