Chapter 3: Problem 62
Looking ahead: Derivative of \(x^{n}\) Use the definition \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) to find \(f^{\prime}(x)\) for the follow functions. a. \(f(x)=x^{2}\) b. \(f(x)=x^{3}\) c. \(f(x)=x^{4}\) d. Based on your answers to parts (a)-(c), propose a form
Short Answer
Expert verified
Answer: The general form for the derivative of \(x^n\) is \(f'(x) = nx^{n-1}\).
Step by step solution
01
a. Finding f'(x) for f(x) = x^2
This exercise asks for finding the derivative of \(f(x) = x^2\). Let's compute the derivative using the definition:
\(f'(x) = \lim_{h \rightarrow 0} \frac{(x+h)^2 - x^2}{h}\)
Now we simplify the expression and apply the limit.
02
Expansion of the expression in the limit
Expand the expression \((x+h)^2\):
\(f'(x) = \lim_{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - x^2}{h}\)
03
Simplify the expression in the limit
Next, we cancel the \(x^2\) terms in the numerator and simplify the remaining expression:
\(f'(x) = \lim_{h \rightarrow 0} \frac{2xh + h^2}{h}\)
04
Factor out h
Factor out \(h\) from the expression inside the limit:
\(f'(x) = \lim_{h \rightarrow 0} h\frac{2x + h}{h}\)
05
Cancel h
Cancel the \(h\) terms:
\(f'(x) = \lim_{h \rightarrow 0} (2x + h)\)
06
Apply the limit
Finally, apply the limit as \(h \rightarrow 0\):
\(f'(x) = 2x\)
So, the derivative of \(f(x) = x^2\) is \(f'(x) = 2x\).
07
b. Finding f'(x) for f(x) = x^3
Now we need to find the derivative of \(f(x)=x^{3}\). Use the definition of the derivative:
\(f'(x) = \lim_{h \rightarrow 0} \frac{(x+h)^3 - x^3}{h}\)
As before, we'll expand, simplify, and apply the limit.
08
Expansion of the expression in the limit
Expand the expression \((x+h)^3\) using the binomial theorem or simply by multiplying \((x+h)\) by itself three times:
\(f'(x) = \lim_{h \rightarrow 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}\)
09
Simplify the expression in the limit
Cancel the \(x^3\) terms in the numerator and simplify the remaining expression:
\(f'(x) = \lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}\)
10
Factor out h
Factor out \(h\) from the expression inside the limit:
\(f'(x) = \lim_{h \rightarrow 0} h\frac{3x^2 + 3xh + h^2}{h}\)
11
Cancel h
Cancel the \(h\) terms:
\(f'(x) = \lim_{h \rightarrow 0} (3x^2 + 3xh + h^2)\)
12
Apply the limit
Finally, apply the limit as \(h \rightarrow 0\):
\(f'(x) = 3x^2\)
So, the derivative of \(f(x) = x^3\) is \(f'(x) = 3x^2\).
13
c. Finding f'(x) for f(x) = x^4
Finally, we need to find the derivative of \(f(x)=x^{4}\) using the same procedure as before:
\(f'(x) = \lim_{h \rightarrow 0} \frac{(x+h)^4 - x^4}{h}\)
Repeat the steps of expansion, simplification, factoring out, cancelation, and applying the limit, just as in parts (a) and (b).
Following the same pattern, the derivative of \(f(x) = x^4\) will be \(f'(x) = 4x^3\).
14
d. Proposing a general form for the derivative of x^n
Based on the results from parts (a), (b), and (c), we consistently see that the derivative of \(x^n\) (where n is a positive integer) takes the general form:
\(f'(x) = nx^{n-1}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
limit of a function
The concept of a "limit of a function" is fundamental in calculus and forms the basis for defining derivatives. When we talk about the limit of a function, we are interested in what value a function approaches as the input gets closer to a certain point. In the context of derivatives, particularly using the definition:
To make this clear, consider \( f(x) = x^2 \). By substituting into the limit definition, we calculate:
- \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
To make this clear, consider \( f(x) = x^2 \). By substituting into the limit definition, we calculate:
- Expand \( (x+h)^2 = x^2 + 2xh + h^2 \)
- Simplify to \( 2x + h \)
- As \( h \to 0, 2x + h \to 2x \)
binomial theorem
The "binomial theorem" is a powerful tool used in algebra to expand expressions that are raised to a power. This is crucial in simplifying expressions within calculus, especially when finding derivatives using limits. The binomial theorem states that for any positive integer \( n \):
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]The theorem allows us to expand expressions such as \( (x+h)^3 \) or \( (x+h)^4 \), turning them into a sum of terms. For instance, \( (x+h)^3 \) can be expanded to:
- \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\)
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]The theorem allows us to expand expressions such as \( (x+h)^3 \) or \( (x+h)^4 \), turning them into a sum of terms. For instance, \( (x+h)^3 \) can be expanded to:
- \( x^3 + 3x^2h + 3xh^2 + h^3 \)
power rule for derivatives
The "power rule for derivatives" is a simple yet extremely useful shortcut in differential calculus. It states that if you have a function \( f(x) = x^n \), where \( n \) is a constant, the derivative is:
- \( f'(x) = nx^{n-1} \)
- For \( x^2 \), derivative is \( 2x \),
- For \( x^3 \), derivative is \( 3x^2 \),
- For \( x^4 \), derivative is \( 4x^3 \)