Problem 18 - Brief Applied Calculus

Question

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\sqrt{x^{6}+3 x-1} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{6x^5 + 3}{2\sqrt{x^6 + 3x - 1}} \).

Step 1: Understanding the Function

The function given is \( f(x) = \sqrt{x^6 + 3x - 1} \). This can be rewritten using the exponent notation as \( f(x) = (x^6 + 3x - 1)^{1/2} \). We will apply the Generalized Power Rule for derivatives.

Step 2: Applying the Generalized Power Rule

According to the Generalized Power Rule, if \( y = (u(x))^n \), then the derivative \( \frac{dy}{dx} = n \cdot (u(x))^{n-1} \cdot u'(x) \). Here, \( n = 1/2 \) and \( u(x) = x^6 + 3x - 1 \). We will differentiate \( u(x) \) first.

Step 3: Differentiating the Inside Function

Compute the derivative of \( u(x) = x^6 + 3x - 1 \). Differentiating each term gives \( u'(x) = 6x^5 + 3 \).

Step 4: Combining Results

Now substitute \( n = 1/2 \), \( u(x) = x^6 + 3x - 1 \), and \( u'(x) = 6x^5 + 3 \) into the Generalized Power Rule. The derivative is: \[ f'(x) = \frac{1}{2} \cdot (x^6 + 3x - 1)^{-1/2} \cdot (6x^5 + 3) \].

Step 5: Simplifying the Derivative

Finally, simplify the expression by multiplying and organizing the terms: \[ f'(x) = \frac{6x^5 + 3}{2\sqrt{x^6 + 3x - 1}} \].

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In simpler terms, it tells us the slope of a function at any given point. This is crucial for understanding the behavior of functions, especially when dealing with curves or non-linear expressions. When you find the derivative of a function, you are essentially finding the rate of change. To find a derivative, we use various rules and formulas. In our problem, we use the Generalized Power Rule, a handy tool when dealing with powers of functions. By applying these rules, we can find derivatives efficiently and systematically. Remember these key points about derivatives:

• They measure instantaneous rates of change.
• They can be used to determine maximum and minimum points on a graph.
• They are essential for solving complex equations and modeling real-world phenomena.

Exponent Notation

Exponent notation is a way of representing repeated multiplication of a number by itself. It simplifies expressions, making them easier to read and work with, especially when dealing with large powers. In mathematical terms, exponent notation is expressed as \( a^n \), where \( a \) is the base and \( n \) is the exponent.In the context of our exercise, exponent notation helps us rewrite the square root function in a more convenient form. By expressing the function as \( (x^6 + 3x - 1)^{1/2} \), it becomes easier to apply the Generalized Power Rule. This transformation is essential, as many differentiation techniques rely on well-understood formats like this.Key points about exponent notation:

• It compresses multiplication into a compact form.
• It's crucial for transforming functions into a form suitable for differentiation.
• Understanding when to apply it aids in algebraic manipulation.

Differentiation

Differentiation is the process of finding the derivative of a function. It involves applying specific rules and techniques to discover how the function behaves at every point over its domain. Differentiation is at the heart of calculus and is used extensively in fields like physics, engineering, and economics. In our exercise, differentiation is performed by applying the Generalized Power Rule. This rule enables us to easily take the derivative of a composite function raised to a power. This is particularly useful when the function inside the power has its own variables to differentiate. Steps for effective differentiation:

• Identify the form of the function.
• Apply rules like the Generalized Power Rule strategically.
• Combine derivatives of individual terms when necessary.

Differentiation doesn't just stop at finding derivatives. Understanding the implications of these derivatives is key to solving many practical problems.