Problem 23 - Calculus of a Single Variable

Question

Finding the Derivative by the Limit Process In Exercises \(11-24,\) find the derivative of the function by the limit process. $$ f(x)=\sqrt{x+4} $$

Step-by-Step Solution

Answer

The derivative of \(f(x) = \sqrt{x+4}\) is \(f'(x) = \frac{1}{2\sqrt{x+4}}\).

Step 1: Writing the limit of the derivative

The limit definition of a derivative expresses the derivative of \(f(x)\) as: \[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]. This will be the starting point.

Step 2: Substituting \(f(x)\) into the derivative limit

Substitute \(f(x)\) and \(f(x + h)\) into the derivative limit equation: \[f'(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x+h+4} - \sqrt{x+4}}{h}\]

Step 3: Rationalizing the numerator

The difference of two square roots in the numerator is difficult to work with. To simplify, multiply the fraction by the conjugate of the numerator, which is \(\sqrt{x+h+4} + \sqrt{x+4}\), over itself. This will not change the value of the fraction, since any number (except zero) divided by itself equals 1. \[f'(x) = \lim_{h \rightarrow 0} \frac{(\sqrt{x+h+4} - \sqrt{x+4})(\sqrt{x+h+4} + \sqrt{x+4})}{h(\sqrt{x+h+4} + \sqrt{x+4})}\] This simplifies the numerator by squaring the roots and canceling terms: \[f'(x) = \lim_{h \rightarrow 0} \frac{(x+h+4) - (x+4)}{h(\sqrt{x+h+4} + \sqrt{x+4})} = \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x+h+4} + \sqrt{x+4})}\] We can further simplify by canceling \(h\) from the numerator and denominator: \[f'(x) = \lim_{h \rightarrow 0} \frac{1}{\sqrt{x+h+4} + \sqrt{x+4}}\]

Step 4: Evaluating the limit

Now the limit can be easily calculated by substituting \(h = 0\) into the expression: \[f'(x) = \frac{1}{\sqrt{x+4} + \sqrt{x+4}} = \frac{1}{2\sqrt{x+4}}\]

Key Concepts

Limit Definition of Derivative

When we talk about derivatives, we're often referring to the rate at which a function is changing at any given point. One of the most fundamental ways to find this rate is through the limit definition of the derivative. This definition tells us how derivatives are calculated using the formula: \[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]Here, the idea is to see what happens as \(h\) approaches zero, which is essentially the gap between two close points on the graph of the function.

• \(f(x+h)\) represents the function evaluated at a point slightly to the right of \(x\).
• \(f(x)\) is the function evaluated directly at \(x\).
• \(h\) is a small gap that gets infinitely closer to zero.

This elegant formula captures the notion of "instantaneous rate of change" or how fast \(f(x)\) is changing at the precise point \(x\). To apply this, we substitute our specific function, in this case \(f(x) = \sqrt{x+4}\), into this limit formula, which sets us up for the next steps.

Rationalizing the Numerator

One of the challenges of applying the derivative limit formula to functions involving square roots is the complex numerator that arises. When you subtract two square roots, it can be almost impossible to simplify directly. This is where the trick of rationalizing the numerator is employed. To rationalize, we multiply the numerator and the denominator by the conjugate of the numerator. If your numerator is \(\sqrt{a} - \sqrt{b}\), its conjugate is \(\sqrt{a} + \sqrt{b}\). This multiplication exploits the identity \((a-b)(a+b) = a^2 - b^2\), eliminating the square roots:

• \((\sqrt{x+h+4} - \sqrt{x+4})(\sqrt{x+h+4} + \sqrt{x+4}) = (x+h+4) - (x+4)\)
• This simplifies to just \(h\), making the expression much easier to handle.

This strategy is key when dealing with functions involving square roots, as it simplifies the expression substantially, allowing us to cancel terms and solve the limit effectively.

Evaluating Limits

Once we have rationalized the numerator and simplified our expression, our next task is to actually evaluate the limit. After simplification, you'll find that many factors cancel each other out, leaving a much simpler expression. In our case, after canceling \(h\) from the numerator and denominator, we are left with: \[f'(x) = \lim_{h \rightarrow 0} \frac{1}{\sqrt{x+h+4} + \sqrt{x+4}}\] By substituting \(h=0\) directly into this expression, we can evaluate the limit:

• As \(h\) approaches 0, \(\sqrt{x+h+4}\) approaches \(\sqrt{x+4}\).
• The expression simplifies to \(\frac{1}{2\sqrt{x+4}}\).

This final step gives us the derivative of the function, providing the rate of change at any point \(x\), and completes the limit process for derivative calculation.

Square Roots in Calculus

Square roots often appear in calculus problems, particularly when you're dealing with functions involving growth, decay, or any form with a radical expression. They can make differentiation tricky, but with the right approach, these can be simplified effectively. In calculus, square roots can pose issues because: - They create complex fractions when subtracted. - They often come up in physics and engineering problems requiring precise calculations. Rationalizing the numerator is a powerful method to wield against the confusion caused by square roots. It allows us to convert complicated, hard-to-handle expressions into simpler terms that are far easier to differentiate. By understanding how square roots operate within the calculus framework, and using strategies like rationalization, you'll stand a better chance at solving these types of calculus challenges without fear.