Problem 4 - Single Variable Calculus

Question

Use the Product Rule to find the derivative of each function. \(f(x)=\frac{e^{x}}{x}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \frac{e^x}{x}\) is \(f'(x) = \frac{e^x(x - 1)}{x^2}\).

Step 1: Rewrite the Function as a Product

Let \(g(x) = e^x\) and \(h(x) = x^{-1}\). Then, the function \(f(x)\) can be written as a product of these two functions: \(f(x) = g(x)\cdot h(x) = e^x \cdot x^{-1}\). Now, we need to find the derivatives of \(g(x)\) and \(h(x)\).

Step 2: Find the Derivatives of g(x) and h(x)

To find the derivatives, we use the following rules of differentiation: 1. The derivative of \(e^x\) is \(e^x\). 2. The derivative of \(x^n\) is \(n \cdot x^{n-1}\). Using these rules, we have: \(g'(x) = e^x\). \(h'(x) = -1 \cdot x^{-2} = -x^{-2}\). Now we will use the product rule to find the derivative of \(f(x)\).

Step 3: Apply the Product Rule

According to the product rule, the derivative of a product of two functions is given by: \((g\cdot h)'(x) = g'(x)\cdot h(x) + h'(x)\cdot g(x)\). Substituting the expressions and derivatives from Steps 1 and 2, we get: \(f'(x) = g'(x)\cdot h(x) + h'(x)\cdot g(x) = e^x\cdot x^{-1} + (-x^{-2})\cdot e^x\).

Step 4: Simplify the Expression

Finally, we can simplify the expression a bit: \(f'(x) = \frac{e^x}{x} - \frac{e^x}{x^2} = \frac{e^x(x - 1)}{x^2}\). So, the derivative of the given function is: \(f'(x) = \frac{e^x(x - 1)}{x^2}\).

Key Concepts

Derivative of e^x

The exponential function with base e, denoted as e^x, is quite special in calculus due to its unique property — the rate at which it grows is directly proportional to its value. This property leads to a compelling result: the derivative of e^x with respect to x is simply e^x itself. This self-deriving character simplifies calculations involving exponential growth or decay problems. For example, in our exercise, the derivative of g(x) = e^x is g'(x) = e^x, which is an initial step before applying the Product Rule for differentiation.

When you encounter e^x in a differentiation problem, you can confidently state its derivative without further transformation. This straightforward rule should be a go-to tool in your calculus toolkit, easing the process of more complex differentiation challenges.

Differentiation Rules

In calculus, different functions require various rules for differentiation. Among the most frequently used are the Power Rule, the Product Rule, and the Chain Rule. The Power Rule states that the derivative of x^n (where n is a real number) is n*x^(n-1). For instance, applying the Power Rule to h(x) = x^-1 in our exercise yields h'(x) = -1 * x^-2.

The Product Rule, which is critical for our exercise, is applied when a function is the product of two other functions. It states that the derivative of the product g(x)*h(x) is g'(x)*h(x) + g(x)*h'(x). Understanding and applying these rules correctly can demystify the process of differentiation and is integral to solving calculus problems effectively.

Function Simplification

Function simplification in calculus involves reducing expressions to a more manageable form, often making it easier to understand or further manipulate the mathematical expressions. Simplifying functions can involve factoring, combining like terms, or canceling common factors. In our exercise's context, the final derivative f'(x) = e^x*x^-1 - e^x*x^-2, is simplified by combining terms over a common denominator, resulting in f'(x) = (e^x*(x - 1))/x^2.

Such simplification not only makes the final answer neater but also facilitates further computations, should the problem require it. For instance, evaluating the derivative at a particular x-value or integrating the simplified derivative can be much more straightforward when the function is in its simplified form. Acquiring the skill to streamline functions systematically will enhance your calculus proficiency.