Problem 25 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions. $$v=\frac{1+x-4 \sqrt{x}}{x}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dv}{dx} = -\frac{1}{x^2} + \frac{2}{x^{3/2}} \).

Step 1: Simplifying the Function

First, simplify the given function by dividing each term in the numerator by the denominator. The function is given as \( v = \frac{1 + x - 4\sqrt{x}}{x} \). By dividing each term by \( x \), we have: \[ v = \frac{1}{x} + \frac{x}{x} - \frac{4\sqrt{x}}{x} \]This simplifies to:\[ v = x^{-1} + 1 - 4x^{-1/2} \]

Step 2: Applying the Derivative Rules

Apply the power rule to each term separately. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).- For \( x^{-1} \), the derivative is \( -x^{-2} \).- For \( 1 \), the derivative is \( 0 \) since the derivative of a constant is zero.- For \( -4x^{-1/2} \), the derivative is \( -4 \times \left( -\frac{1}{2} \right) x^{-3/2} = 2x^{-3/2} \).

Step 3: Writing the Final Derivative

Combine all the derivatives of the terms to express the final derivative of the function. The derivative of the given function is:\[ \frac{dv}{dx} = -x^{-2} + 0 + 2x^{-3/2} \]This simplifies to:\[ \frac{dv}{dx} = -\frac{1}{x^2} + \frac{2}{x^{3/2}} \].

Key Concepts

Understanding the Power Rule

The power rule is a fundamental tool in calculus, particularly when finding derivatives. It makes handling polynomial functions and functions expressed in terms of exponents much simpler. The rule states that if you have a function of the form \( f(x) = x^n \), where \( n \) is a real number, the derivative of this function is \( f'(x) = n \cdot x^{n-1} \). This is because you're essentially reducing the power by one and then multiplying it by the original power itself.

When using the power rule, here are key things to remember:

• Apply the derivative to each term in the sum of functions separately.
• Remember, the derivative of any constant is zero.
• Licensed to work with both positive and negative exponents, and even fractional exponents!

In our example of finding the derivative of \( v = x^{-1} + 1 - 4x^{-1/2} \), you apply the power rule to each term. For example, the derivative of \( x^{-1} \) becomes \( -x^{-2} \). The middle \( 1 \) vanishes, as its derivative is \( 0 \), and \( -4x^{-1/2} \) becomes \( 2x^{-3/2} \). This demonstrates the versatility and strength of the power rule in calculus problem-solving.

Simplifying Functions Before Differentiation

Simplifying functions might not just simplify the math, but also drastically reduces calculation errors later on. Before applying any calculus rules like the power rule, it's beneficial to simplify the given function as much as possible.

Consider how we simplified the exercise function \( v = \frac{1 + x - 4\sqrt{x}}{x} \). By breaking it down, each term in the numerator was divided by the denominator individually:

• \( \frac{1}{x} \) becomes \( x^{-1} \).
• \( \frac{x}{x} \) simplifies to simply \( 1 \).
• \( \frac{4\sqrt{x}}{x} \) transforms into \( 4x^{-1/2} \).

This transformation results in \( v = x^{-1} + 1 - 4x^{-1/2} \), an expression far easier to differentiate. This step is vital, as calculus can get complex quickly. Simplification is thus a great preparatory stage, setting up the function nicely for smooth differentiation.

Basics of Calculus Problem-Solving

Calculus problem-solving often involves a clear understanding of the concepts and orderly steps to arrive at a solution. The procedure typically starts with understanding the problem, simplifying if necessary, applying the correct calculus rule, and then solving.

In the derivative problem presented, after simplification, each function component was separately differentiated before combining them back together. It emphasizes the importance of stepwise breakdown:

• Simplify: Break down complex functions to make each term manageable.
• Select the Right Rule: Use the power rule here, but other rules might be suitable based on function nature.
• Apply the Rule: Operate on each term, adhering to rule specifics.
• Combine and Simplify: After deriving, combine parts to form a final solution, simplifying if necessary.

Through this process, beginner calculus topics become more approachable. Insight into each step demystifies complex math and aids in developing effective problem-solving skills. With practice, these steps become second nature in solving various calculus problems efficiently.