Problem 53 - Calculus

Question

Estimate the value of \(f^{\prime}(1)\) by substitutang small values for \(h .\) Then use algebra to find \(f^{\prime}(1)\) exactly.$$f(x)=\sqrt{x}$$

Step-by-Step Solution

Answer

The exact value of \(f^{\prime}(1) = 0.5\).

Step 1: 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 \(x = 1\).

Step 2: 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 \(x = 1\).

Step 3: 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. \(0.499\).

Step 4: 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, \(f^{\prime}(x) = \frac{1}{2}x^{-1/2}\).

Step 5: 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 \(h\).

Step 6: 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 \(f^{\prime}(1) = 0.5\).

Key Concepts

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:

• \( f^{\prime}(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)

Therefore, the process involves finding the difference \( f(a+h) - f(a) \) and dividing it by \( h \), and then calculating the limit as \( h \) approaches zero.

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:

• \( \frac{d}{dx}(x^{n}) = nx^{n-1} \)

This rule simplifies the differentiation process significantly and is particularly useful for functions that can be expressed as powers of \( x \).

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:

• \( f^{\prime}(x) = \frac{1}{2}x^{-1/2} \)

This powerful rule lets you find the derivative function quickly, making it easier to evaluate at any point, such as \( x = 1 \).

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:

• 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.

For differentiation, you may convert a square root function like \( \sqrt{x} \) into its exponential form, \( x^{1/2} \).
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.