Problem 25 - Calculus for Biology and Medicine

Question

Use the formal definition to find the derivative of \(y=\sqrt{x}\) at \(x=2\)

Step-by-Step Solution

Answer

The derivative of \(y=\sqrt{x}\) at \(x=2\) is \(\frac{\sqrt{2}}{2}\).

Step 1: Identify the Function and Point

We start with the function \(y = \sqrt{x}\) and need to find the derivative at \(x = 2\). This means determining the slope of the tangent line to the curve at this particular point.

Step 2: Apply the Definition of the Derivative

The derivative of a function \(f(x)\) at a point \(x = a\) is given by the limit: \(f'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h}\). For our function \(f(x) = \sqrt{x}\) and \(a = 2\), this becomes \(f'(2) = \lim_{{h \to 0}} \frac{{\sqrt{2+h} - \sqrt{2}}}{h}\).

Step 3: Rationalize the Numerator

To evaluate the limit, we often rationalize the numerator. Multiply and divide by the conjugate: \[\lim_{{h \to 0}} \frac{{\sqrt{2+h} - \sqrt{2}}}{h} \times \frac{{\sqrt{2+h} + \sqrt{2}}}{\sqrt{2+h} + \sqrt{2}} = \lim_{{h \to 0}} \frac{(\sqrt{2+h})^2 - (\sqrt{2})^2}{h(\sqrt{2+h} + \sqrt{2})}.\]

Step 4: Simplify the Expression

Simplify the expression: \((\sqrt{2+h})^2 - (\sqrt{2})^2 = (2+h) - 2 = h.\) So the limit simplifies to \[\lim_{{h \to 0}} \frac{h}{h(\sqrt{2+h} + \sqrt{2})} = \lim_{{h \to 0}} \frac{1}{\sqrt{2+h} + \sqrt{2}}.\]

Step 5: Evaluate the Limit

As \(h \to 0\), the expression \(\sqrt{2+h}\) approaches \(\sqrt{2}\). Thus, the limit becomes: \[\lim_{{h \to 0}} \frac{1}{\sqrt{2+h} + \sqrt{2}} = \frac{1}{2\sqrt{2}}.\] Simplifying, we find that \(f'(2) = \frac{\sqrt{2}}{2}.\)

Key Concepts

Formal Definition of Derivative

The formal definition of a derivative is a crucial concept in calculus that helps us understand the rate at which a function is changing at any point. In simple terms, the derivative of a function at a point gives us the slope of the tangent line at that particular point.

To find the derivative of a function at a specific point, such as for the function \( y = \sqrt{x} \) at \( x = 2 \), we use the definition:

• Identify the function \( f(x) = \sqrt{x} \).
• Find its derivative at the point \( x = a \).

The formula for the derivative at a point \( a \) is given by:\[f'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h}.\]Here, the term \( f(a+h) \) involves evaluating the function close to \( a \), effectively capturing how the function \( f(x) \) changes in a very tiny neighborhood around \( a \).

This limit process is not only a foundation for derivative computation but also opens doors to understanding how functions behave locally. Understanding this concept is essential for analyzing curves and optimizing functions.

Limits

Limits can seem like a complex concept, but they are foundational in calculus. They describe how a function behaves as the input approaches a certain value. This concept is especially vital when we need to understand situations where straightforward computation might not directly apply due to indeterminacies.

In the case of finding the derivative of \( y = \sqrt{x} \) at \( x = 2 \), the limit notation appears as part of the derivative formula:\[\lim_{{h \to 0}} \frac{{\sqrt{2+h} - \sqrt{2}}}{h}.\]The purpose of the limit here is to see what happens to the expression as \( h \) gets closer and closer to zero.

It's a mechanism that allows us to evaluate expressions that become undefined or ambiguous for direct substitution.

• Helps handle cases where the expression could be undefined, like dividing by zero.
• Crucial for solving real-world problems involving instantaneous rates and slopes.

So, when you see a limit, think of it as a tool that lets calculus "zoom in" on a point to find what's happening in its immediate vicinity.

Rationalizing the Numerator

Rationalizing is a technique often used in calculus to simplify expressions to make limits easier to evaluate. This approach is particularly beneficial when dealing with square roots in the numerator, a common scenario when finding derivatives of functions involving roots.

In the derivative calculation for \( y = \sqrt{x} \) at \( x = 2 \), we encounter:\[ \lim_{{h \to 0}} \frac{{\sqrt{2+h} - \sqrt{2}}}{h}\]Evaluating this directly is tricky due to the square root terms. By rationalizing the numerator, we eliminate those troublesome radicals:

• Multiply the numerator and denominator by the conjugate: \( \sqrt{2+h} + \sqrt{2} \).
• This generates a difference of squares: \((\sqrt{2+h})^2 - (\sqrt{2})^2 = h \).
• The expression simplifies, allowing the limit to be evaluated without encountering division by zero.

Rationalizing aids in isolating and cancelling terms that obscure the limit's true value. In this case, it reveals a clean and solvable form, so we can accurately determine the derivative at a specific point. This technique highlights why algebraic manipulation is so powerful in calculus, enabling the solution of limits that initially appear resistant to straightforward approaches.