Chapter 2: Problem 20
Find a formula for the derivative using the power rule. Confirm it using difference quotients. $$l(x)=1 / x^{2}$$
Short Answer
Expert verified
The derivative of \( l(x) = \frac{1}{x^2} \) is \( l'(x) = -\frac{2}{x^3} \).
Step by step solution
01
Rewrite Using Exponents
First, rewrite the given function in a form suitable for using the power rule. The function is currently in the form \( l(x) = \frac{1}{x^2} \). This can be rewritten as \( l(x) = x^{-2} \).
02
Apply the Power Rule
The power rule states that if \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \). In this case, \( n = -2 \). Applying the power rule, we get: \( l'(x) = -2 \cdot x^{-2-1} = -2 \cdot x^{-3} = -\frac{2}{x^3} \).
03
Set Up the Difference Quotient
The difference quotient for the derivative is \( \lim_{{h \to 0}} \frac{l(x+h) - l(x)}{h} \). Substitute \( l(x) = x^{-2} \) into this formula, giving us \( \lim_{{h \to 0}} \frac{(x+h)^{-2} - x^{-2}}{h} \).
04
Simplify the Difference Quotient
Expand the expression: \( (x+h)^{-2} = \frac{1}{(x+h)^2} \) and \( x^{-2} = \frac{1}{x^2} \). The difference quotient becomes \( \lim_{{h \to 0}} \frac{\left(\frac{1}{(x+h)^2} - \frac{1}{x^2}\right)}{h} \). Simplify the numerator: \( \frac{x^2 - (x+h)^2}{x^2(x+h)^2} \).
05
Algebraically Simplify
Continue simplifying the expression \( x^2 - (x+h)^2 = x^2 - (x^2 + 2xh + h^2) \) which simplifies to \(-2xh - h^2\). Thus, the difference quotient becomes \( \lim_{{h \to 0}} \frac{-2xh - h^2}{h \cdot x^2(x+h)^2} \).
06
Evaluate the Limit
Factor out \( h \) in the numerator: \( \lim_{{h \to 0}} \frac{h(-2x - h)}{h \cdot x^2(x+h)^2} \). Cancel \( h \): \( \lim_{{h \to 0}} \frac{-2x - h}{x^2(x+h)^2} \). Substitute \( h = 0 \) to find the limit: \( \frac{-2x}{x^2(x)^2} = -\frac{2}{x^3} \).
07
Confirm the Derivative
The result from the difference quotient is \( -\frac{2}{x^3} \), which matches the result obtained from applying the power rule. Thus, the derivative of \( l(x) = \frac{1}{x^2} \) is confirmed as \( l'(x) = -\frac{2}{x^3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule
The Power Rule is a fundamental concept in calculus that simplifies the process of finding the derivative of polynomial functions. When we have a function in the form of \( f(x) = x^n \), the Power Rule tells us that the derivative is \( f'(x) = n \cdot x^{n-1} \).
This means you multiply the variable's exponent by the coefficient and reduce the exponent by one. It's a quick and reliable way to find derivatives, especially when dealing with terms like \( x^3 \) or \( x^{-2} \).
In the given problem, we had the function \( l(x) = \frac{1}{x^2} \), which we rewrote as \( l(x) = x^{-2} \). Applying the Power Rule here:
This means you multiply the variable's exponent by the coefficient and reduce the exponent by one. It's a quick and reliable way to find derivatives, especially when dealing with terms like \( x^3 \) or \( x^{-2} \).
- Start by identifying the exponent of the term.
- Apply the rule: multiply by the exponent.
- Decrease the exponent by one and adjust the term.
In the given problem, we had the function \( l(x) = \frac{1}{x^2} \), which we rewrote as \( l(x) = x^{-2} \). Applying the Power Rule here:
- We determine \( n = -2 \).
- The derivative then becomes \( l'(x) = -2 \cdot x^{-3} = -\frac{2}{x^3} \).
Difference Quotient
The Difference Quotient is a crucial concept that represents the average rate of change of a function over an interval. More formally, it's the expression used to define the derivative:
\[ \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
This formula is foundational in calculus because it leads to the derivative by letting the interval \( h \) approach zero.
In this exercise, we applied the Difference Quotient to verify the derivative found using the Power Rule:
The Difference Quotient not only finds the derivative independently, but also corroborates methods like the Power Rule. Its thorough step-by-step approach builds a deeper understanding of how functions change, ensuring accuracy in calculus.
\[ \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
This formula is foundational in calculus because it leads to the derivative by letting the interval \( h \) approach zero.
In this exercise, we applied the Difference Quotient to verify the derivative found using the Power Rule:
- We substituted \( l(x) = x^{-2} \) into the difference quotient.
- Simplifications were made leading to the expression \( \lim_{{h \to 0}} \frac{(x+h)^{-2} - x^{-2}}{h} \).
- Next, algebraic manipulations helped reduce this to \( \lim_{{h \to 0}} \frac{-2x-h}{x^2(x+h)^2} \).
- Finally, substituting \( h = 0 \), we obtained the confirmed derivative, \( -\frac{2}{x^3} \).
The Difference Quotient not only finds the derivative independently, but also corroborates methods like the Power Rule. Its thorough step-by-step approach builds a deeper understanding of how functions change, ensuring accuracy in calculus.
Limit Definition
The Limit Definition in calculus is the underlying principle for understanding derivatives. A limit represents the value that a function approaches as the input approaches a certain point. In terms of derivatives, the limit is used to define the concept of the derivative at a specific point.
The formula is:
\[ \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
This formula captures the idea of a tangent line to a curve. As \( h \to 0 \), the secant line approaches the tangent line, giving us the derivative, which is the instantaneous rate of change.
Using this for the function \( l(x) = \frac{1}{x^2} \), we:
The limit definition is essential because it provides a rigorous foundation for derivatives, ensuring that any derivative found accurately reflects the function's behavior as inputs vary closely around a given point.
The formula is:
\[ \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \]
This formula captures the idea of a tangent line to a curve. As \( h \to 0 \), the secant line approaches the tangent line, giving us the derivative, which is the instantaneous rate of change.
Using this for the function \( l(x) = \frac{1}{x^2} \), we:
- Expanded the expression to \( \lim_{{h \to 0}} \frac{1/(x+h)^2 - 1/x^2}{h} \).
- Performed algebraic simplifications and factored where possible.
- Evaluated the resulting simplified expression as \( h \to 0 \).
- This process confirmed \( l'(x) = -\frac{2}{x^3} \).
The limit definition is essential because it provides a rigorous foundation for derivatives, ensuring that any derivative found accurately reflects the function's behavior as inputs vary closely around a given point.