Chapter 2: Problem 55
Estimate the value of \(f^{\prime}(1)\) by substitutang small values for \(h .\) Then use algebra to find \(f^{\prime}(1)\) exactly.$$f(x)=2 x+x^{-1}$$
Short Answer
Expert verified
The exact value of \(f'(1)\) is 1.
Step by step solution
01
Understand the Problem
We need to estimate the derivative \(f'(1)\) of the function \(f(x) = 2x + x^{-1}\) using small values of \(h\), and then use algebraic methods to find the exact value of the derivative at \(x = 1\).
02
Use Definition of Derivative for Estimation
The derivative \(f'(x)\) at any point \(x\) can be estimated using the difference quotient definition:\[f'(x) \approx \frac{f(x + h) - f(x)}{h}\]Let \(x = 1\), so we estimate \(f'(1)\) by substituting small increments \(h\):\[f'(1) \approx \frac{f(1 + h) - f(1)}{h}\]
03
Calculate Function Values
First, calculate \(f(1)\):\[f(1) = 2(1) + (1)^{-1} = 2 + 1 = 3\]Now, calculate \(f(1 + h)\):\[f(1 + h) = 2(1 + h) + (1 + h)^{-1}\]
04
Compute Approximation
Plug the values from the previous step into the difference quotient:\[f'(1) \approx \frac{2(1 + h) + (1 + h)^{-1} - 3}{h}\]Simplify this expression for very small values of \(h\) (e.g., \(h = 0.1, 0.01, 0.001\)) to get estimated values of \(f'(1)\). Make sure to compute \((1+h)^{-1}\) using specific \(h\) values.
05
Simplify Expression to Find Exact Derivative
Find the derivative \(f'(x)\) using algebraic differentiation. \[f(x) = 2x + x^{-1}\]Differentiate term by term:- The derivative of \(2x\) is \(2\).- The derivative of \(x^{-1}\) is \(-x^{-2}\) (using the power rule \((x^n)' = nx^{n-1}\)).Thus,\[f'(x) = 2 - x^{-2}\]
06
Find Exact Value of f'(1)
Substitute \(x = 1\) into the derivative:\[f'(1) = 2 - (1)^{-2} = 2 - 1 = 1\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference Quotient
The difference quotient is a technique used to estimate the derivative of a function at a particular point. It involves evaluating the function at a point slightly offset by a small increment, often denoted as \(h\). This method is based on the definition of the derivative as \(f'(x) \approx \frac{f(x + h) - f(x)}{h}\).
- When \(h\) is very small, the difference quotient provides a good approximation of the derivative.
- This approach is especially useful for estimating derivatives when algebraic differentiation is complicated or impractical.
Algebraic Differentiation
Algebraic differentiation is the process of finding the exact derivative of a function using algebraic techniques. Unlike the estimation given by the difference quotient, algebraic differentiation provides an exact symbolic result.
- The derivative of \(2x\) is 2 because the slope of a line described by \(2x\) is constant.- The derivative of \(x^{-1}\) requires the power rule, yielding \(-x^{-2}\).
Thus, the exact derivative is \(f'(x) = 2 - x^{-2}\), which gives us the precise rate of change at any point \(x\), including \(x = 1\), leading to \(f'(1) = 1\).
- It is often performed by applying rules for differentiation to find a new function that represents the rate of change of the original function.
- This method eliminates the approximation error associated with numerical methods, making it highly accurate and preferable when an exact value is required.
- The derivative of \(2x\) is 2 because the slope of a line described by \(2x\) is constant.- The derivative of \(x^{-1}\) requires the power rule, yielding \(-x^{-2}\).
Thus, the exact derivative is \(f'(x) = 2 - x^{-2}\), which gives us the precise rate of change at any point \(x\), including \(x = 1\), leading to \(f'(1) = 1\).
Power Rule
The power rule is a fundamental principle in calculus used for differentiating functions where variables are raised to a power. This rule makes deriving any polynomial function straightforward.
In mathematical terms, if you have a function in the form \(x^n\), its derivative using the power rule is \((x^n)' = nx^{n-1}\). This indicates that you multiply the variable's exponent by its coefficient and then reduce the exponent by one.
In mathematical terms, if you have a function in the form \(x^n\), its derivative using the power rule is \((x^n)' = nx^{n-1}\). This indicates that you multiply the variable's exponent by its coefficient and then reduce the exponent by one.
- This rule is efficient and widely applicable across various problems involving polynomials.
- It simplifies the process of differentiation, converting a potentially complex task into an easy procedure.