Problem 18 - Calculus Single Variable

Question

Use the Product Rule to compute the derivative of the given expression with respect to \(x\). (In each of Exercises 7,8,14,16, and 18, do not avoid using the Product Rule by first expanding the product.) $$ (1 / x) \cdot\left(5 x^{3}+4 / x\right) $$

Step-by-Step Solution

Answer

The derivative is \(10x - \frac{8}{x^3}\).

Step 1: Identify the Functions

In the expression \( \frac{1}{x} \cdot (5x^3 + \frac{4}{x}) \), we have two functions that we need to work with. Let \( u(x) = \frac{1}{x} \) and \( v(x) = 5x^3 + \frac{4}{x} \). Our objective is to differentiate the product of these two functions, \( u(x) \cdot v(x) \).

Step 2: Derivative of the First Function

Calculate the derivative of \( u(x) = \frac{1}{x} \). The derivative \( u'(x) \) can be found using the power rule for negative exponents, yielding \( u'(x) = \frac{d}{dx}(x^{-1}) = -x^{-2} \).

Step 3: Derivative of the Second Function

Now, differentiate \( v(x) = 5x^3 + \frac{4}{x} \). For \( v(x) \), the derivative is \( v'(x) = \frac{d}{dx}(5x^3) + \frac{d}{dx}(4x^{-1}) = 15x^2 - 4x^{-2} \).

Step 4: Apply the Product Rule

According to the product rule, \((uv)' = u'v + uv'\). Substitute the derivatives and original functions: \((uv)' = (-x^{-2})(5x^3 + \frac{4}{x}) + (\frac{1}{x})(15x^2 - 4x^{-2})\).

Step 5: Simplify the Expression

Distribute each term: \(-x^{-2} \cdot 5x^3 = -5x \) and \(-x^{-2} \cdot \frac{4}{x} = -\frac{4}{x^3}\). Also, \( \frac{1}{x} \cdot 15x^2 = 15x \) and \( \frac{1}{x} \cdot -4x^{-2} = -\frac{4}{x^3} \). Sum these terms to get the overall derivative: \(15x - 5x - \frac{4}{x^3} - \frac{4}{x^3} = 10x - \frac{8}{x^3}\).

Key Concepts

Understanding Derivatives

Before tackling the product of two functions using the Product Rule, it's crucial to understand what a derivative is. In simple terms, the derivative of a function is a way to measure how that function changes. Imagine driving a car: the derivative tells you the speed of the car, or how fast the distance (your function) is changing over time (your variable). In mathematics, the derivative gives us the slope of the function at any point, showing how quickly it rises or falls.
To find the derivative, we look at tiny changes in the function and how those changes relate to changes in the input. Calculating derivatives can involve several rules and techniques, one of which is the Product Rule when dealing with multiplication of functions.

The Power Rule

When you hear about differentiating functions, you'll often come across the Power Rule. The Power Rule is a fundamental tool used to calculate derivatives when a function is raised to a power, like the expressions you might see like \(x^n\). This rule significantly simplifies the process of finding the derivative of powers of \(x\).
The Power Rule states that if you have a function \(f(x) = x^n\), its derivative \(f'(x)\) is found by multiplying the power by the coefficient and then subtracting one from the power. Therefore, \(f'(x) = nx^{n-1}\).

• Example: For \(f(x) = x^3\), the derivative \(f'(x) = 3x^{2}\).
• This simple rule is also applied to negative and fractional exponents.

Learning to apply the Power Rule seamlessly opens the doors to mastering more complex rules like the Product Rule.

Working with Functions

Functions play a central role in calculus and analysis. A function is simply a rule that takes an input and gives out exactly one output. Think of it like a machine where you insert a number, and out comes another number after processing.

• For example, the function \(f(x) = x^2\) takes an input \(x\) and outputs the square of \(x\).
• Different types of functions, such as polynomial, exponential, or trigonometric functions have different behaviors and rules for differentiation.

Knowing the kind of function you are dealing with is essential in applying the correct differentiation techniques. In the original exercise, you encountered a product of functions that required the use of the Product Rule for accurate differentiation.

Simplification through Algebra

After applying the Product Rule, the resulting computational expression often needs simplification. Simplification in mathematics is the process of making an expression easier to work with or more concise without changing its value.
For differential calculus, it ensures that the final derivative expression is in its simplest and most easily interpretable form. This involves combining like terms and reducing fractions wherever possible.

• For instance, if you arrive at \( -5x + 15x\), you can simplify it down to \(10x\).
• Similarly, terms like \( -\frac{4}{x^3} \) are simplified by recognizing common bases and combining them.

Practicing simplification helps prevent errors and enhances problem-solving efficiency, making complex calculus problems much more manageable.