Problem 20
Question
Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\log \left(1+x^{2}\right) \text { at } a=0 $$
Step-by-Step Solution
Verified Answer
The linear approximation of \(f(x)\) at \(a=0\) is \(L(x) = 0\).
1Step 1: Understand the Problem
We need to approximate the function \(f(x) = \log(1+x^2)\) at \(a = 0\) using the linear approximation formula: \(L(x) = f(a) + f'(a)(x-a)\). This will involve calculating both \(f(a)\) and \(f'(a)\).
2Step 2: Calculate \(f(a)\)
Substitute \(a = 0\) into the function \(f(x)\). We have \(f(0) = \log(1+0^2) = \log(1) = 0\).
3Step 3: Find the Derivative \(f'(x)\)
Differentiate \(f(x) = \log(1+x^2)\) with respect to \(x\). Using the chain rule, \(f'(x) = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}\).
4Step 4: Evaluate \(f'(a)\)
Substitute \(a = 0\) into the derivative \(f'(x)\), giving \(f'(0) = \frac{2(0)}{1+0^2} = \frac{0}{1} = 0\).
5Step 5: Substitute into Linear Approximation Formula
Plug \(f(a)\) and \(f'(a)\) into the linear approximation formula: \(L(x) = 0 + 0 \cdot (x-0) = 0\). Therefore, \(L(x) = 0\).
Key Concepts
CalculusDerivativeChain Rule
Calculus
Calculus is a branch of mathematics that studies how quantities change and how they come together in various forms. One of the fundamental aspects of calculus is using functions to represent problems.
The idea is to use a mathematical function to outline how changes in a variable can affect a whole system. In the given exercise, calculus helps us approach approximation through a powerful tool in mathematics: linear approximation.
The purpose of linear approximation is to find a linear equation that is closest to a function at a given point. In this case, the function we are examining is \( f(x) = \log(1+x^2) \), and our task is to approximate it at \( a = 0 \). By breaking it down into manageable steps, calculus allows us to make accurate predictions about a function's behavior near that point.
The idea is to use a mathematical function to outline how changes in a variable can affect a whole system. In the given exercise, calculus helps us approach approximation through a powerful tool in mathematics: linear approximation.
The purpose of linear approximation is to find a linear equation that is closest to a function at a given point. In this case, the function we are examining is \( f(x) = \log(1+x^2) \), and our task is to approximate it at \( a = 0 \). By breaking it down into manageable steps, calculus allows us to make accurate predictions about a function's behavior near that point.
Derivative
In calculus, the derivative of a function measures how the function's output changes as the input changes. It provides us with the rate of change or the slope at any given point.
For linear approximation, understanding the derivative is essential because the slope of the tangent line at a point becomes a key part of the approximation formula. In this exercise, identifying the derivative of \( f(x) = \log(1+x^2) \) allows us to build the linear model.
To find the derivative \( f'(x) \), we use calculus techniques such as the chain rule to handle more complex functions. Once the derivative is calculated, \( f'(x) = \frac{2x}{1+x^2} \), we evaluated it at \( x=0 \) to see the slope at that point, which is \( 0 \). This zero slope is crucial in understanding why the linear approximation results in a constant value.
For linear approximation, understanding the derivative is essential because the slope of the tangent line at a point becomes a key part of the approximation formula. In this exercise, identifying the derivative of \( f(x) = \log(1+x^2) \) allows us to build the linear model.
To find the derivative \( f'(x) \), we use calculus techniques such as the chain rule to handle more complex functions. Once the derivative is calculated, \( f'(x) = \frac{2x}{1+x^2} \), we evaluated it at \( x=0 \) to see the slope at that point, which is \( 0 \). This zero slope is crucial in understanding why the linear approximation results in a constant value.
Chain Rule
The chain rule is a method used in calculus for differentiating composite functions, which are functions made up of two or more functions combined together.
Imagine you need the derivative of a function like \( f(x) = \log(1+x^2) \). The chain rule helps by breaking down the differentiation process into simpler parts. First, think of \( 1+x^2 \) as an inside function and \( \log \, u \) as an outside function where \( u = 1 + x^2 \).
To apply the chain rule, differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function. Therefore, \( f'(x) = \frac{1}{1+x^2} \times 2x \), leading us to the expression \( \frac{2x}{1+x^2} \). This application is a staple of calculus, often employed when handling more convoluted functions. Using the chain rule makes complicated derivatives easier to tackle and integrate into solutions.
Imagine you need the derivative of a function like \( f(x) = \log(1+x^2) \). The chain rule helps by breaking down the differentiation process into simpler parts. First, think of \( 1+x^2 \) as an inside function and \( \log \, u \) as an outside function where \( u = 1 + x^2 \).
To apply the chain rule, differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function. Therefore, \( f'(x) = \frac{1}{1+x^2} \times 2x \), leading us to the expression \( \frac{2x}{1+x^2} \). This application is a staple of calculus, often employed when handling more convoluted functions. Using the chain rule makes complicated derivatives easier to tackle and integrate into solutions.
Other exercises in this chapter
Problem 19
Apply the product rule to find the tangent line, in slope-intercept form, of \(y=f(x)\) at the specified point. $$ f(x)=4\left(2 x^{4}+3 x\right)\left(4-2 x^{2}
View solution Problem 19
Differentiate the functions given in Problems with respect to the independent variable. $$ f(x)=20 x^{3}-4 x^{6}+9 x^{8} $$
View solution Problem 20
Compute \(f(c+h)-f(c)\) at the indicated point. $$ f(x)=\frac{1}{x} ; c=-2 $$
View solution Problem 20
Differentiate the functions with respect to the independent variable. \(h(s)=\frac{2(3-s)^{2}}{s^{2}+(7 s-1)^{2}}\)
View solution