Problem 18 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Show that \(\frac{d}{d x} \sqrt{x+8}=\frac{1}{2 \sqrt{x+8}}\) using the limit de nition of derivative. You ll use different versions of the de nition in parts (a) and (b). In both cases it will be necessary to rationalize the numerator in order to evaluate the limit. (a) \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) (b) \(f^{\prime}(x)=\lim _{b \rightarrow x} \frac{f(b)-f(x)}{b-x}\)

Step-by-Step Solution

Answer

The derivative of \(\sqrt{x+8}\) is \(\frac{1}{2 \sqrt{x+8}}\), as proven using both forms of the limit definition of the derivative.

Step 1: Understand the problem

The problem is to verify the derivative of \(\sqrt{x+8}\), using the definition of derivative twice – once for each case of the limit definition of derivative.

Step 2: Solve for (a)

First apply the limit definition of derivative for (a): \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\). \n Substituting the function \(f(x) = \sqrt{x+8}\) in the above definition gives: \[ f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x+h+8} - \sqrt{x+8}}{h} \] To rationalize the numerator, multiply and divide by the conjugate: \[ f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x+h+8} - \sqrt{x+8}}{h} \times \frac{\sqrt{x+h+8} + \sqrt{x+8}}{\sqrt{x+h+8} + \sqrt{x+8}} = \lim_{h \rightarrow 0} \frac{(x+h+8) - (x+8)}{h(\sqrt{x+h+8} + \sqrt{x+8})} = \frac{1}{2 \sqrt{x+8}} \] as \(h \rightarrow 0\).

Step 3: Solve for (b)

Now apply the limit definition of derivative for (b): \(f^{\prime}(x) = \lim_{b \rightarrow x} \frac{f(b)-f(x)}{b-x}\). Substituting the function \(f(x) = \sqrt{x+8}\) in the above definition as well gives: \[ f^{\prime}(x) = \lim_{b \rightarrow x} \frac{\sqrt{b+8} - \sqrt{x+8}}{b-x} \] To rationalize the numerator, multiply and divide by the conjugate: \[ f^{\prime}(x) = \lim_{b \rightarrow x} \frac{\sqrt{b+8} - \sqrt{x+8}}{b-x} \times \frac{\sqrt{b+8} + \sqrt{x+8}}{\sqrt{b+8} + \sqrt{x+8}} = \lim_{b \rightarrow x} \frac{(b+8) - (x+8)}{(b-x)(\sqrt{b+8} + \sqrt{x+8})} = \frac{1}{2 \sqrt{x+8}} \] as \(b \rightarrow x\).

Step 4: Finalize the solution

For both scenarios (a) and (b), the derivative of \(\sqrt{x+8}\) is \(\frac{1}{2 \sqrt{x+8}}\). This is now successfully proven using the limit definition of the derivative in both the scenarios provided.

Key Concepts

Limit Definition of Derivative

The limit definition of the derivative is a fundamental mathematical concept that allows us to transition from a difference quotient to an instantaneous rate of change – the derivative. The most common version of this definition involves a little helper, known as the "difference quotient," which is given by:

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

Here, \( h \) is a small increment added to \( x \), and this formula calculates how \( f(x) \) changes as \( x \) increases by \( h \). As \( h \) approaches zero, the difference quotient becomes the derivative. It's like measuring a car's speed on a perfectly straight, infinitely small road.
The alternative but equivalent form involves a variable \( b \) approaching \( x \) as follows:

• \( f^{\prime}(x) = \lim_{b \rightarrow x} \frac{f(b) - f(x)}{b-x} \)

This setup is perfect for examples where a specific value, \( b \), trends towards \( x \), and we're interested in the derivative at that precise point. Both forms achieve the same goal, finding the instant slope at any given \( x \). Clarifying this, let's move to rationalizing the numerator, which helps evaluate these limits.

Rationalizing the Numerator

Rationalizing the numerator is a clever algebraic trick to simplify expressions, especially important in calculus when dealing with roots. When you encounter a square root in a fraction, multiplying both the numerator and denominator by the conjugate can eliminate the root from the numerator. This process is vital when applying the limit definition of a derivative.
For example, if you have an expression like:

• \( \frac{\sqrt{x+h+8} - \sqrt{x+8}}{h} \)

We multiply both the numerator and denominator by the conjugate of the numerator:

• \( \frac{\sqrt{x+h+8} - \sqrt{x+8}}{h} \times \frac{\sqrt{x+h+8} + \sqrt{x+8}}{\sqrt{x+h+8} + \sqrt{x+8}} \)

This simplifies to:

• \( \frac{(x+h+8) - (x+8)}{h (\sqrt{x+h+8} + \sqrt{x+8})} \)

The result? The pesky \( h \) cancels out, allowing us to comfortably take the limit as \( h \rightarrow 0 \) without division by zero. This method can significantly streamline finding derivatives in problems involving roots.

Fundamental Theorem of Calculus

So why are derivatives a big deal? Because of the fundamental theorem of calculus that bridges the worlds of differentiation (derivatives) and integration. This theorem states that if you have a continuous function on an interval and an antiderivative — a function whose derivative gives back the original function — then the definite integral of that function over the interval gives you the net area under the curve.
In simple terms, the fundamental theorem explains how a derivative unwinds an integral and vice versa. It's like a math seesaw. You differentiate to find how fast something is changing at an instant, and you integrate to find the overall total change over a period.
This connection also underscores why derivatives, found using limits as in the given problem, are crucial. Understanding and being able to prove derivative values sets the stage for more advanced calculus topics, such as optimizing functions or computing areas and volumes.