Chapter 23: Problem 7
Find the derivative of each of the functions by using the definition. $$y=x^{2}-1$$
Short Answer
Expert verified
The derivative of the function \(y = x^2 - 1\) is \(f'(x) = 2x\).
Step by step solution
01
Understand the Definition of the Derivative
The derivative of a function \(y = f(x)\) at a point \(x\) is given by the limit: \[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]For the function \(y = x^2 - 1\), we need to compute this limit to find the derivative.
02
Substitute into the Derivative Formula
We start by finding \(f(x+h)\). For our function \(f(x) = x^2 - 1\), we have: \[ f(x+h) = (x+h)^2 - 1 \]Now, plug the expressions into the derivative formula:\[ f'(x) = \lim_{{h \to 0}} \frac{(x+h)^2 - 1 - (x^2 - 1)}{h} \]
03
Simplify the Expression
First, expand \((x+h)^2\) using the binomial formula:\[ (x+h)^2 = x^2 + 2xh + h^2 \]Substituting back, we simplify:\[ f'(x) = \lim_{{h \to 0}} \frac{x^2 + 2xh + h^2 - 1 - x^2 + 1}{h} \]Cancel the \(x^2\) and \(-1\) terms:\[ f'(x) = \lim_{{h \to 0}} \frac{2xh + h^2}{h} \]
04
Factor and Cancel the \(h\)
Factor \(h\) out of the numerator:\[ f'(x) = \lim_{{h \to 0}} \frac{h(2x + h)}{h} \]Since \(h eq 0\) (except in the limit, which doesn't affect division), cancel \(h\):\[ f'(x) = \lim_{{h \to 0}} (2x + h) \]
05
Evaluate the Limit
Now, evaluate the limit by substituting \(h = 0\):\[ f'(x) = 2x + 0 = 2x \]Thus, the derivative of the function \(y = x^2 - 1\) is \(f'(x) = 2x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, a derivative measures how a function changes as its input changes. It is essentially the slope of the function at a particular point. Let's take the example of the function given:
For the function \( y = x^2 - 1 \), the derivative, denoted \( f'(x) \), shows us the instantaneous rate of change of the function with respect to \( x \). To find the derivative, you apply the definition by calculating the limit:
- Substitute \( f(x+h) \) into the formula.- Simplify the expression.- Factor and cancel terms if possible.- Evaluate the limit as \( h \) approaches 0.
By working through these steps, for \( y = x^2 - 1 \), the result is \( f'(x) = 2x \). This means for any point on the parabola \( y = x^2 - 1 \), the slope of the tangent line at that point is \( 2x \).
For the function \( y = x^2 - 1 \), the derivative, denoted \( f'(x) \), shows us the instantaneous rate of change of the function with respect to \( x \). To find the derivative, you apply the definition by calculating the limit:
- Substitute \( f(x+h) \) into the formula.- Simplify the expression.- Factor and cancel terms if possible.- Evaluate the limit as \( h \) approaches 0.
By working through these steps, for \( y = x^2 - 1 \), the result is \( f'(x) = 2x \). This means for any point on the parabola \( y = x^2 - 1 \), the slope of the tangent line at that point is \( 2x \).
Limit
Limits are foundational in calculus, providing the basis for defining derivatives, integrals, and continuity. A limit captures the behavior of a function as its input approaches a particular value.
When taking the derivative, we compute:\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]This limit represents the idea of considering the slope as \( h \), a small increase in \( x \), approaches zero.
It is important to note that as \( h \) approaches zero, we never actually allow \( h \) to be zero since the expression \( \frac{f(x+h) - f(x)}{h} \) would be undefined. Instead, we explore values of \( h \) that are exceedingly small to perceive the trend or behavior of the function.
When taking the derivative, we compute:\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]This limit represents the idea of considering the slope as \( h \), a small increase in \( x \), approaches zero.
It is important to note that as \( h \) approaches zero, we never actually allow \( h \) to be zero since the expression \( \frac{f(x+h) - f(x)}{h} \) would be undefined. Instead, we explore values of \( h \) that are exceedingly small to perceive the trend or behavior of the function.
Binomial Formula
The binomial formula is a mathematical tool that is often used to expand expressions raised to a power. In calculus, it assists in the computation of derivatives, particularly when dealing with polynomial functions.
For example, consider the expression \( (x+h)^2 \), which appears in the process of finding derivatives:- By expanding, the expression becomes \( x^2 + 2xh + h^2 \).- This expansion is derived using the binomial theorem, which states:\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]In our case, \( a = x \), \( b = h \), and \( n = 2 \). The expanded form helps us simplify our expressions when calculating limits and derivatives.
For example, consider the expression \( (x+h)^2 \), which appears in the process of finding derivatives:- By expanding, the expression becomes \( x^2 + 2xh + h^2 \).- This expansion is derived using the binomial theorem, which states:\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]In our case, \( a = x \), \( b = h \), and \( n = 2 \). The expanded form helps us simplify our expressions when calculating limits and derivatives.
Function Analysis
Function analysis involves examining the behavior, properties, and structure of mathematical functions. It's a broad field in calculus that includes understanding how a function behaves at various points and intervals.
For the function \( y = x^2 - 1 \), we already know:- It is a quadratic function represented by a parabola.- The vertex lies at (0, -1), serving as the minimum point.
Analyzing its derivative, \( f'(x) = 2x \), provides insights into:- **Slope**: The slope is positive when \( x > 0 \) and negative when \( x < 0 \), indicating the parabola opens upwards.- **Intervals of Increase/Decrease**: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).Function analysis aids in deeper understanding, not only predicting function behavior but also optimizing and solving real-world problems involving motion, growth, and more.
For the function \( y = x^2 - 1 \), we already know:- It is a quadratic function represented by a parabola.- The vertex lies at (0, -1), serving as the minimum point.
Analyzing its derivative, \( f'(x) = 2x \), provides insights into:- **Slope**: The slope is positive when \( x > 0 \) and negative when \( x < 0 \), indicating the parabola opens upwards.- **Intervals of Increase/Decrease**: The function is increasing for \( x > 0 \) and decreasing for \( x < 0 \).Function analysis aids in deeper understanding, not only predicting function behavior but also optimizing and solving real-world problems involving motion, growth, and more.