Problem 59 - Calculus Volume 1

Question

For the following exercises, use the definition of a derivative to find \(f^{\prime}(x)\) . $$f(x)=\sqrt{2 x}$$

Step-by-Step Solution

Answer

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

Step 1: Recall the Definition of a Derivative

The derivative of a function \( f(x) \) at any point \( x \) is given by the limit: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]

Step 2: Substitute the Function into the Definition

Substitute \( f(x) = \sqrt{2x} \) into the derivative definition:\[ f'(x) = \lim_{{h \to 0}} \frac{{\sqrt{2(x+h)} - \sqrt{2x}}}{h} \]

Step 3: Simplify the Expression Using Conjugates

Multiply and divide by the conjugate to simplify:\[ f'(x) = \lim_{{h \to 0}} \frac{(\sqrt{2(x+h)} - \sqrt{2x})(\sqrt{2(x+h)} + \sqrt{2x})}{h (\sqrt{2(x+h)} + \sqrt{2x})} \]This simplifies to: \[ f'(x) = \lim_{{h \to 0}} \frac{2(x+h) - 2x}{h(\sqrt{2(x+h)} + \sqrt{2x})} = \lim_{{h \to 0}} \frac{2h}{h(\sqrt{2(x+h)} + \sqrt{2x})} \]

Step 4: Simplify Further and Take the Limit

Cancel \( h \) from the numerator and denominator:\[ f'(x) = \lim_{{h \to 0}} \frac{2}{\sqrt{2(x+h)} + \sqrt{2x}} \]As \( h \to 0 \), \( \sqrt{2(x+h)} \to \sqrt{2x} \), so:\[ f'(x) = \frac{2}{2\sqrt{2x}} \]

Step 5: Simplify Final Expression

The expression simplifies to:\[ f'(x) = \frac{1}{\sqrt{2x}} \]

Key Concepts

Limit Definition of Derivative

The concept of a derivative is central in calculus. At its core, it represents how a function changes at a particular point. The limit definition is a precise way to define this rate of change. For a function \( f(x) \), its derivative, denoted \( f'(x) \), is defined as:

• \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]

This formula captures the essence of calculus: calculating limits to understand change. This definition allows us to find the slope of the tangent line to any curve at any given point. For students, it's important to realize that \( h \) approaches zero so that we are precisely measuring the change at that exact point. By applying this to various functions, you practice the essential skill of finding derivatives.

Simplifying Expressions

Simplification plays a key role when using the limit definition of derivatives. In our example, after substituting the function into the derivative formula, we face a radical expression that needs simplification. To handle such expressions, employing the conjugate is an effective method.

• Multiply and divide the expression by its conjugate: \(\sqrt{2(x+h)} + \sqrt{2x}\).
• This technique helps clear the radicals from the numerator, making it easier to simplify.

Upon multiplying by the conjugate, you will notice that the numerator becomes a difference of squares, which simplifies considerably. It's crucial to cancel terms properly, particularly \( h \) in the numerator and denominator, to correctly compute the limit. Understanding these steps enhances your ability to simplify complex expressions and solve derivatives more efficiently.

Derivatives of Radical Functions

Radical functions, such as those involving square roots, can seem difficult to differentiate at first. However, the procedure becomes manageable by using the limit definition and simplifying the expression correctly.

• Start by expressing the function appropriately in the derivative formula.
• Use algebraic techniques, like multiplying by the conjugate, to handle any radicals present.
• Apply limit properties to evaluate as \( h \to 0 \).

For \( f(x) = \sqrt{2x} \), the derivative comes out to be \( f'(x) = \frac{1}{\sqrt{2x}} \). This expresses how the slope of this radical function decreases as \( x \) increases, highlighting its behavior graphically. Mastering derivatives of these functions enables deeper insights into their geometry and real-world applications in fields like physics and engineering.