Question

Assuming the limit exists, the definition of the derivative \(f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\) implies that if h is small, then an approximation to \(f^{\prime}(a)\) is given by $$ f^{\prime}(a) \approx \frac{f(a+h)-f(a)}{h} $$If \(h>0,\) then this approximation is called a forward difference quotient; if \(h<0,\) it is a backward difference quotient. As shown in the following exercises, these formulas are used to approximate \(f^{\prime}\) at a point when \(f\) is a complicated function or when \(f\) is represented by a set of data points. Let \(f(x)=\sqrt{x}\) a. Find the exact value of \(f^{\prime}(4)\) b. Show that \(f^{\prime}(4) \approx \frac{f(4+h)-f(4)}{h}=\frac{\sqrt{4+h}-2}{h}\) c. Complete columns 2 and 5 of the following table and describe how \(\frac{\sqrt{4+h}-2}{h}\) behaves as \(h\) approaches 0 d. The accuracy of an approximation is measured by error \(=|\) exact value \(-\) approximate value \(|\) Use the exact value of \(f^{\prime}(4)\) in part (a) to complete columns 3 and 6 in the table. Describe the behavior of the errors as \(h\) approaches 0.

Short answer

Answer: As h approaches 0 from both positive and negative values, the forward and backward difference quotients get closer to the exact value of the derivative, which is \(\frac{1}{4}\). This means that the approximations become more accurate as the spacing between points gets smaller. Additionally, both errors from positive and negative values of h will get closer to 0 as h goes to 0, indicating the approximations become more accurate and the asymmetry in the errors cancels out.

Step-by-step solution

Finding the exact value of f'(4)

To find the exact value of \(f'(4)\), we need to first find the derivative of the function \(f(x) = \sqrt{x}\). Since \(\sqrt{x} = x^{1/2}\), we can simply apply the power rule for derivatives: \(f'(x) = \frac{1}{2}x^{-1/2}\) Now that we have \(f'(x)\), we can find the exact value by plugging in \(x = 4\): \(f'(4) = \frac{1}{2}(4)^{-1/2} = \frac{1}{2}(2)^{-1} = \frac{1}{4}\) So, the exact value of \(f'(4)\) is \(\frac{1}{4}\).

Showing the approximation

We already know that $$ f^{\prime}(a) \approx \frac{f(a+h)-f(a)}{h} $$ Plugging in \(f(x)=\sqrt{x}\) and \(a=4\), we get: $$ f^{\prime}(4) \approx \frac{\sqrt{4+h}-\sqrt{4}}{h} = \frac{\sqrt{4+h}-2}{h} $$

Completing columns 2 and 5 of the table

The second column of the table represents the forward differences for a range of \(h\) values. For each \(h\) value, we need to compute the corresponding forward difference using the approximation we derived earlier: \(\frac{\sqrt{4+h}-2}{h}\). Calculating this for the different \(h\) values will complete column 2. Similarly, column 5 represents the backward differences, and we must calculate the same approximation with negative \(h\) values to complete column 5.

Describing how the approximation behaves as h approaches 0

As \(h\) gets closer and closer to 0 from both positive and negative values, the approximation \(\frac{\sqrt{4+h}-2}{h}\) gets closer and closer to the exact value of \(f'(4)\), which is \(\frac{1}{4}\). This demonstrates that the forward and backward difference quotients both approach the derivative value as \(h\) goes to 0.

Using the exact value of f'(4) to complete columns 3 and 6 in the table and describe the behavior of the errors

The error is defined as the absolute difference between the exact value (\(f'(4)\)) and the approximate value. Since we know the exact value of \(f'(4)\) is \(\frac{1}{4}\), we can compute the error based on the approximations we found in columns 2 and 5. For each value of \(h\), compute the value of error as follows: $$ \text{error} = \left|\frac{1}{4} - \frac{\sqrt{4+h}-2}{h}\right| $$ Complete columns 3 and 6 by computing the error for each corresponding entry in columns 2 and 5.

Describing the behavior of the errors as h approaches 0

As \(h\) approaches 0 from both positive and negative values, the error values (columns 3 and 6) will get closer and closer to 0. This indicates that the forward and backward difference quotients become more accurate as the spacing between points gets smaller. The positive and negative values of \(h\) will effectively cancel out any asymmetry present in the errors.

Key concepts

Difference Quotient

In calculus, the difference quotient is a crucial concept used to estimate the derivative of a function at a specific point. It essentially provides a means to measure how a function changes as its input changes by a small amount. This is mathematically expressed as:

• Forward difference quotient for positive h: \( \frac{f(a+h) - f(a)}{h} \)
• Backward difference quotient for negative h: \( \frac{f(a-h) - f(a)}{h} \)

The forward difference quotient is used when \( h \) is greater than zero, providing an approximation by looking forward from point \( a \). Conversely, the backward difference quotient uses negative \( h \) to approximate by looking backward. These quotients approach the derivative as \( h \) gets closer to zero, capturing the function's instantaneous rate of change. During exercises, you typically calculate these quotients with specific increments to see how they converge to the actual derivative value, like we did with \( \sqrt{x} \) at \( x = 4 \).

Derivative Approximation

The general idea behind derivative approximation is to use simple calculations to find how close we can get to the actual derivative. In our exercise, we use \( \frac{\sqrt{4+h} - 2}{h} \) as an approximation for \( f'(4) \), where \( f(x) = \sqrt{x} \). This approximation mirrors the forward difference quotient. By plugging in smaller and smaller values for \( h \), you can observe how the approximation asymptotically approaches the real derivative value.
To further understand this, try several different values of \( h \), both positive and negative, to observe the convergence pattern. As \( h \) nears zero, the approximation error decreases, and the computed result gets closer to the true derivative value, demonstrating the practical utility of this method in analyzing more complex functions or datasets where direct computation of the derivative might be challenging.

Power Rule

The power rule is a simple yet powerful tool in calculus for finding the derivative of a function expressed as a power of \( x \). If you have a function \( f(x) = x^n \), its derivative \( f'(x) \) is given by \( nx^{n-1} \). This rule is straightforward to apply, making it a favorite among students learning derivatives for the first time.
For example, when finding the derivative of \( f(x) = \sqrt{x} = x^{1/2} \), we apply the power rule:

• The exponent, \( n = \frac{1}{2} \), is multiplied by the base \( x \) raised to the power of \( n-1 \), which simplifies to \( \frac{1}{2}x^{-1/2} \).

The power rule allows you to quickly determine that \( f'(x) = \frac{1}{2}x^{-1/2} \), and thus find \( f'(4) = \frac{1}{4} \) by substituting \( x = 4 \). The ease of use and application efficiency makes the power rule a fundamental part of calculus.

Error Analysis

Error analysis in derivative approximation involves assessing how well our approximate derivatives compare to the exact derivative. The error is calculated as the absolute difference between the exact derivative value and the approximation. In our example exercise, the exact value of \( f'(4) \) is \( \frac{1}{4} \), which we then compared with different approximations, such as \( \frac{\sqrt{4+h}-2}{h} \).

• For each \( h \) value used, the error is computed as: \( |\frac{1}{4} - \frac{\sqrt{4+h}-2}{h}| \).

As you decrease the size of \( h \), these errors should reduce and become closer to zero, indicating better approximation accuracy. This reduction shows how error analysis aids in verifying the effectiveness and reliability of your derivative approximations, providing confidence in their use as \( h \) becomes infinitesimally small.