Chapter 2: Problem 53
Estimate the value of \(f^{\prime}(1)\) by substitutang small values for \(h .\) Then use algebra to find \(f^{\prime}(1)\) exactly.$$f(x)=\sqrt{x}$$
Short Answer
Expert verified
The exact value of \(f^{\prime}(1) = 0.5\).
Step by step solution
01
Understanding the Problem
We are tasked with estimating the derivative of the function \(f(x) = \sqrt{x}\) at \(x = 1\) using small values of \(h\), and finding the exact derivative algebraically at \(x = 1\).
02
Using the Definition of Derivative
The derivative of a function \(f(x)\) at a point \(x = a\) is given by the limit \(f^{\prime}(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\). We will substitute small values for \(h\) to estimate this limit at \(x = 1\).
03
Substituting Small Values for \(h\)
Let's estimate \(f^{\prime}(1)\) using small values of \(h\). For instance, try \(h = 0.1\), \(h = 0.01\), and \(h = 0.001\). Calculate \(\frac{\sqrt{1+h} - 1}{h}\) for each.- For \(h = 0.1\), \(\frac{\sqrt{1.1} - 1}{0.1}\) approx. \(0.487\).- For \(h = 0.01\), \(\frac{\sqrt{1.01} - 1}{0.01}\) approx. \(0.499\).- For \(h = 0.001\), \(\frac{\sqrt{1.001} - 1}{0.001}\) approx. \(0.499\).
04
Algebraic Calculation of the Derivative
Now calculate the derivative \(f^{\prime}(x)\) of \(f(x) = \sqrt{x}\) using algebra. Rewrite \(\sqrt{x}\) as \(x^{1/2}\). Differentiate using the power rule: \(\frac{d}{dx}(x^{n}) = nx^{n-1}\). So, \(f^{\prime}(x) = \frac{1}{2}x^{-1/2}\).
05
Finding the Exact Value at \(x = 1\)
Substitute \(x = 1\) into the derivative: \(f^{\prime}(1) = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}\). This confirms the estimates from small values of \(h\).
06
Conclusion
The derivative \(f^{\prime}(1) = \frac{1}{2}\) is approximately consistent with our estimates from small \(h\) values. The exact value using calculus is \(f^{\prime}(1) = 0.5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Definition of Derivative
The derivative of a function at a specific point measures how the function's output changes as its input changes, quantifying the function's 'instantaneous rate of change' at a point. One of the fundamental ways to define a derivative is through the limit definition.
The limit definition of the derivative gives us a precise mathematical way to calculate this rate of change. Specifically, for a function \( f(x) \) at a point \( x = a \):
This is given by:
This process can be estimated by choosing very small values for \( h \). For the exercise, you can use values like \( h = 0.1 \), \( h = 0.01 \), and \( h = 0.001 \), calculating the approximate values. As \( h\) becomes smaller, the estimation becomes more accurate, approaching the exact derivative value.
The limit definition of the derivative gives us a precise mathematical way to calculate this rate of change. Specifically, for a function \( f(x) \) at a point \( x = a \):
This is given by:
- \( f^{\prime}(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
This process can be estimated by choosing very small values for \( h \). For the exercise, you can use values like \( h = 0.1 \), \( h = 0.01 \), and \( h = 0.001 \), calculating the approximate values. As \( h\) becomes smaller, the estimation becomes more accurate, approaching the exact derivative value.
Power Rule
The power rule is a fundamental tool in calculus for finding derivatives. It provides a straightforward way to differentiate functions of the form \( x^n \), where \( n \) is a real number. The power rule states that:
In the original exercise, we can express the function \( f(x) = \sqrt{x} \) in terms of a power, as \( x^{1/2} \).
Using the power rule, the derivative of \( x^{1/2} \) is calculated as follows:
The algebraic method of differentiation often results in the same or more precise outcome compared to the limit definition.
- \( \frac{d}{dx}(x^{n}) = nx^{n-1} \)
In the original exercise, we can express the function \( f(x) = \sqrt{x} \) in terms of a power, as \( x^{1/2} \).
Using the power rule, the derivative of \( x^{1/2} \) is calculated as follows:
- \( f^{\prime}(x) = \frac{1}{2}x^{-1/2} \)
The algebraic method of differentiation often results in the same or more precise outcome compared to the limit definition.
Square Root Function
Square root functions are a type of radical function where the variable \( x \) is under a square root. The general form is \( f(x) = \sqrt{x} \).
These functions are common in mathematics and have unique characteristics and behaviors. Here are a few important points about square root functions:
This makes it easier to differentiate using the power rule.
Once differentiated, it turns into an expression \( \frac{1}{2}x^{-1/2} \), allowing you to find specific rates of change at required points, such as \( x = 1 \). This insight into how square root functions behave not only aids in understanding their derivatives but also provides clarity on how small changes in \( x \) affect the outcome.
These functions are common in mathematics and have unique characteristics and behaviors. Here are a few important points about square root functions:
- They only exist for non-negative inputs since you can't take the square root of a negative number in the real number system.
- The output is always non-negative.
- The function is increasing, meaning it always goes upwards as \( x \) increases.
This makes it easier to differentiate using the power rule.
Once differentiated, it turns into an expression \( \frac{1}{2}x^{-1/2} \), allowing you to find specific rates of change at required points, such as \( x = 1 \). This insight into how square root functions behave not only aids in understanding their derivatives but also provides clarity on how small changes in \( x \) affect the outcome.