Problem 20 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ y=x \cdot 2^{x} $$

Step-by-Step Solution

Answer

The derivative is \(y' = 2^{x}(1 + x \cdot \ln(2))\).

Step 1: Identify the Function

The given function is \(y = x \cdot 2^{x}\). This is the product of two separate functions: \(f(x) = x\) and \(g(x) = 2^{x}\). To find the derivative of their product, we will use the product rule.

Step 2: Recall the Product Rule

The product rule states that if you have a function \(y = u(x) \cdot v(x)\), the derivative \(y'\) is given by \(y' = u'(x)\cdot v(x) + u(x) \cdot v'(x)\). In our case, \(u(x) = x\) and \(v(x) = 2^{x}\).

Step 3: Differentiate Each Function Separately

First, differentiate \(u(x) = x\) which gives \(u'(x) = 1\). Next, differentiate \(v(x) = 2^{x}\). The derivative of \(2^{x}\) is \(v'(x) = 2^{x} \cdot \ln(2)\).

Step 4: Apply the Product Rule

Substitute the derivatives into the product rule formula: \(y' = (1) \cdot 2^{x} + x \cdot 2^{x} \cdot \ln(2)\).

Step 5: Simplify the Derivative

Combine the terms to simplify the expression: \(y' = 2^{x} + x \cdot 2^{x} \cdot \ln(2)\). The derivative \(y'\) can also be factored as \(y' = 2^{x}(1 + x \cdot \ln(2))\).

Key Concepts

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. If you have two functions, let's call them \( u(x) \) and \( v(x) \), and you're trying to find the derivative of their product \( y = u(x) \cdot v(x) \), the product rule provides a formula for this. It states:

• \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

This formula might look a bit intimidating initially, but it's quite straightforward in practice.
Instead of trying to take the derivative of the entire product at once, you treat each component separately:

• First, differentiate \( u(x) \) to get \( u'(x) \).
• Then, differentiate \( v(x) \) to get \( v'(x) \).
• Apply the product rule formula, inserting your derivatives and original functions where they belong.

In simpler terms, the product rule helps break down complex expressions into manageable parts. This makes taking derivatives simpler and more organized.

Exponential Functions

Exponential functions are functions where the variable is in the exponent, such as \( f(x) = a^{x} \), where \( a \) is a constant base and \( x \) is the exponent. These types of functions are crucial in many mathematical and scientific contexts due to their unique properties, such as representing growth, decay, or scaling phenomena.When differentiating exponential functions, a special rule applies. For instance, the derivative of \( a^{x} \) is:

• \( \frac{d}{dx}[a^x] = a^{x} \cdot \ln(a) \)

Here, \( \ln(a) \) is the natural logarithm of the base \( a \). This factor appears because logarithms help to "bring down" exponents, relating them to more easily handled expressions.
Understanding exponential functions and their differentiation is key because these functions often model real-world systems, including population growth, radioactive decay, and financial interest calculations.

Logarithmic Differentiation

Logarithmic differentiation is a clever technique used primarily when dealing with complex functions that are products, quotients, or powers of other functions. It involves taking the natural logarithm of both sides of a function equation and thus leveraging the properties of logarithms to simplify differentiation.Here's how it works:

• Take the natural logarithm of the function \( y = f(x) \), so \( \ln(y) = \ln(f(x)) \).
• Use the properties of logarithms to break down the right-hand side, which can make the differentiation step easier. For example, \( \ln(uv) = \ln(u) + \ln(v) \), which is useful for products.
• Differentiate both sides with respect to \( x \). Remembering that \( \frac{d}{dx}[\ln(y)] = \frac{y'}{y} \), solve for \( y' \).

Logarithmic differentiation simplifies otherwise lengthy differentiation tasks, allowing you to tackle functions that would be cumbersome to differentiate directly. This tool is particularly useful when the function is a product of several terms or when raised to a power.