Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ H(x)=\sqrt{x^{2}+4} $$

Step-by-Step Solution

Answer

The derivative is \( H'(x) = \frac{x}{\sqrt{x^2 + 4}} \).

Step 1: Set Up the Derivative Formula

We need to find the derivative \(H'(x)\) using the limit definition of the derivative. The function is \(H(x) = \sqrt{x^2 + 4}\). According to the limit definition \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\), for our function, it becomes:\[H'(x) = \lim_{h \to 0} \frac{\sqrt{(x+h)^2 + 4} - \sqrt{x^2 + 4}}{h}\]

Step 2: Substitute and Simplify Inside the Limit

We substitute and express \(H(x + h) = \sqrt{(x+h)^2 + 4}\). This gives us:\[H'(x) = \lim_{h \to 0} \frac{\sqrt{x^2 + 2xh + h^2 + 4} - \sqrt{x^2 + 4}}{h}\]

Step 3: Rationalize the Numerator

To simplify the expression, multiply both numerator and denominator by the conjugate of the numerator:\[H'(x) = \lim_{h \to 0} \frac{(\sqrt{x^2 + 2xh + h^2 + 4} - \sqrt{x^2 + 4})(\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})}{h (\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})}\]This simplifies to:\[H'(x) = \lim_{h \to 0} \frac{((x^2 + 2xh + h^2 + 4) - (x^2 + 4))}{h (\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})}\]

Step 4: Further Simplify the Expression

After cancellation in the numerator, we have:\[H'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h (\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})}\]This can be broken down to:\[H'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h (\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})}\]

Step 5: Cancel \(h\) and Take the Limit

Cancel \(h\) in the numerator and denominator:\[H'(x) = \lim_{h \to 0} \frac{2x + h}{\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4}}\]As \(h\) approaches 0, this becomes:\[H'(x) = \frac{2x}{2\sqrt{x^2 + 4}}\]

Step 6: Final Simplification

Simplifying further:\[H'(x) = \frac{x}{\sqrt{x^2 + 4}}\]This gives us the derivative \(H'(x)\) of the function \(H(x) = \sqrt{x^2 + 4}\).

Key Concepts

Limit Definition of Derivative

To find the derivative of a function at a particular point, we use the limit definition of a derivative. This core concept forms the foundation of differential calculus. The limit definition is given as: \( f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \). It represents the idea of finding the slope of the tangent line to the curve at a particular point. In simpler terms, it's the average rate of change of the function as the interval \( h \) becomes infinitely small.

In the context of the function \( H(x) = \sqrt{x^2 + 4} \), our goal is to apply this definition to determine the derivative \( H'(x) \). We substitute the original function and its incremented form \( H(x+h) \) into the formula. This sets the stage for simplifying the expression and rationalizing the numerator.

Rationalizing the Numerator

Rationalizing the numerator is a technique often used to eliminate the square roots in the numerator of a fraction. This process makes it easier to simplify the expression and take limits. When dealing with the expression \( \frac{\sqrt{(x+h)^2 + 4} - \sqrt{x^2 + 4}}{h} \), it becomes challenging to evaluate the limit directly due to the square roots.

By multiplying the numerator and the denominator by the conjugate of the numerator, \( \sqrt{(x+h)^2 + 4} + \sqrt{x^2 + 4} \), we can clear the roots from the numerator. This results in a difference of squares, simplifying it to \( [(x^2 + 2xh + h^2 + 4) - (x^2 + 4)] \) and paving the way for further simplification. Now, the expression becomes more manageable, allowing us to progress in calculating the derivative.

Simplifying Expressions

After rationalizing, the next step is simplifying the resulting expression. We focus on canceling and combining like terms. From the previous result \( \frac{2xh + h^2}{h (\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4})} \), notice how the numerator contains a common \( h \) factor.

Factoring \( h \) out as common allows cancellation with \( h \) in the denominator. This simplification leads to \( \frac{2x + h}{\sqrt{x^2 + 2xh + h^2 + 4} + \sqrt{x^2 + 4}} \). As the limit \( h \to 0 \) is taken, the terms involving \( h \) diminish, and the expression simplifies to a point where we can compute the derivative without ambiguity.

Derivative of Square Root Function

The result of the differentiation process gives us \( H'(x) = \frac{x}{\sqrt{x^2 + 4}} \). This expression is the derivative of the square root function \( H(x) = \sqrt{x^2 + 4} \). Understanding the derivative of a square root function involves recognizing how the rate of change behaves with the non-linear root function.

Square root functions can be tricky due to their non-linearity and the complication introduced by the chain rule when differentiating. The final expression \( \frac{x}{\sqrt{x^2 + 4}} \) tells us about the function's rate of change concerning \( x \). It reflects how each unit change in \( x \) impacts the value of \( H(x) \). This rate of change is essential in understanding the behavior and graph of the function as it grows.