Chapter 2: Problem 70
In each part, compute \(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}\), and then state the formula for \(f^{(n)}\). (a) \(f(x)=1 / x\) (b) \(f(x)=1 / x^{2}\) [Hint: The expression \((-1)^{n}\) has a value of 1 if \(n\) is even and \(-1\) if \(n\) is odd. Use this expression in your answer.]
Short Answer
Step by step solution
Compute the first derivative of f(x) = 1/x
Compute the second derivative of f(x) = 1/x
Compute the third derivative of f(x) = 1/x
Find the formula for the nth derivative of f(x) = 1/x
Compute the first derivative of f(x) = 1/x^2
Compute the second derivative of f(x) = 1/x^2
Compute the third derivative of f(x) = 1/x^2
Find the formula for the nth derivative of f(x) = 1/x^2
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule
In the original exercise, the power rule is applied to functions like \( f(x) = \frac{1}{x} \) and \( f(x) = \frac{1}{x^2} \). These functions are rewritten using negative exponents: \( f(x) = x^{-1} \) and \( f(x) = x^{-2} \).
Using the power rule:
- For \( f(x) = x^{-1} \), the first derivative is \( f'(x) = -1x^{-2} = -\frac{1}{x^2}\).
- For \( f(x) = x^{-2} \), the first derivative is \( f'(x) = -2x^{-3} = -\frac{2}{x^3}\).
Nth Derivative
In the provided solution, after calculating the first few derivatives:
- For \( f(x) = \frac{1}{x} \): We found \( f'(x) = -\frac{1}{x^2} \), \( f''(x) = \frac{2}{x^3} \), and \( f'''(x) = -\frac{6}{x^4} \). The pattern leads to the formula \( f^{(n)}(x) = (-1)^n n! x^{-(n+1)} \).
- For \( f(x) = \frac{1}{x^2} \): We found \( f'(x) = -\frac{2}{x^3} \), \( f''(x) = \frac{6}{x^4} \), and \( f'''(x) = -\frac{24}{x^5} \). The pattern forms the formula \( f^{(n)}(x) = (-1)^n (n+1)! x^{-(n+2)} \).
Derivative Formulas
In the context of the given exercise, we see the use of derivative formulas like:
- Power Rule: As described earlier, useful for functions of the form \( x^n \).
- Factorials in nth Derivatives: For functions such as \( 1/x \) or \( 1/x^2 \), the factorial (\( n! \)) plays a significant role when expressing nth derivatives. This is evident in the formulas derived from the patterns in the solution:
- Alternating Sign: The expression \((-1)^n\) is crucial for alternating the sign of each derivative, depending on whether \( n \) is odd or even. This is observed in the patterns: \( f^{(n)}(x) = (-1)^n n! x^{-(n+1)} \) and \( f^{(n)}(x) = (-1)^n (n+1)! x^{-(n+2)} \).