Chapter 2: Problem 10
Use the definition of the derivative to compute the derivative of the given function. $$g(x)=x^{2}$$
Short Answer
Expert verified
The derivative of \(g(x) = x^2\) is \(g'(x) = 2x\).
Step by step solution
01
Understand the Derivative Definition
The derivative of a function \(g(x)\) can be defined by the limit: \( g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \). Our task is to compute this limit for the function \(g(x) = x^2\).
02
Substitute the Function into the Derivative Formula
Substitute \(g(x) = x^2\) into the limit definition of the derivative: \( g'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} \).
03
Expand the Expression
Expand the expression \((x + h)^2\): \((x + h)^2 = x^2 + 2xh + h^2\). Substitute this back into the limit expression: \( g'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \).
04
Simplify the Expression
Cancel out the \(x^2\) terms in the numerator: \( g'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} \).
05
Factor and Cancel
Factor \(h\) from the numerator: \( g'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h} \). Cancel \(h\) from the numerator and the denominator: \( g'(x) = \lim_{h \to 0} (2x + h) \).
06
Evaluate the Limit
Substitute \(h = 0\) into the expression: \( g'(x) = 2x \). Thus, the derivative of \(g(x) = x^2\) is \(2x\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Definition of Derivative
The limit definition of a derivative is a fundamental concept in calculus. It formally expresses the derivative of a function at a given point. If you think of a function's derivative as its "instantaneous rate of change" or "slope" at a particular moment, you're on the right track! This can be found by calculating a limit. For a function \( g(x) \), the derivative \( g'(x) \) is calculated using the formula:
- \( g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \)
Polynomial Function Differentiation
Differentiating polynomial functions using the limit definition is a common task in calculus. Polynomials, such as \( g(x) = x^2 \), have terms that are powers of \( x \). When differentiating such functions, you'll notice a pattern:
- First, substitute the polynomial into the derivative's limit formula. For example, when \( g(x) = x^2 \), you get: \( g'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \).
- Expand the polynomial expressions where needed. In the \( x^2 \) case, \( (x+h)^2 \) becomes \( x^2 + 2xh + h^2 \).
Simplifying Expressions
Simplifying expressions is a vital skill in calculus, especially when dealing with limits. After substituting and expanding the polynomial, you'll often have a more complex expression that needs simplifying. Consider how we've expanded \( (x + h)^2 \) into \( x^2 + 2xh + h^2 \). The goal is to simplify:
- First, cancel any terms that directly negate each other. For the derivative \( g'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \), the \( x^2 \) terms simplify.
- What remains is \( \frac{2xh + h^2}{h} \), which can be further simplified by factoring out \( h \), giving \( \frac{h(2x + h)}{h} \).
- Finally, cancel out common factors in the numerator and denominator, leaving us with \( 2x + h \).
Evaluating Limits
Evaluating limits is the final step in finding a derivative through its limit definition. After simplifying the expression, you'll need to compute the limit as \( h \to 0 \). This involves direct substitution, as in the example expression \( 2x + h \).
- Since we can't directly substitute when there's an \( h \) in the denominator (which is undefined at \( h = 0 \)), simplification is key.
- Once simplified to \( 2x + h \), substitute \( h = 0 \) to find \( g'(x) = 2x \).