Problem 112 - Calculus Volume 1

Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=(x+2)\left(2 x^{2}-3\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 6x^2 + 8x - 3 \).

Step 1: Identify the Product Rule

The function \( f(x) = (x+2)(2x^2 - 3) \) is a product of two functions: \( u(x) = x + 2 \) and \( v(x) = 2x^2 - 3 \). We will use the product rule for derivatives, which states \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u(x) \)

The function \( u(x) = x + 2 \) is linear. Its derivative \( u'(x) \) is obtained by differentiating term by term: \( u'(x) = 1 \).

Step 3: Differentiate \( v(x) \)

The function \( v(x) = 2x^2 - 3 \) is a polynomial. Apply the power rule: \( v'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3) = 4x \).

Step 4: Apply the Product Rule

Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 1 \cdot (2x^2 - 3) + (x + 2) \cdot 4x \].

Step 5: Simplify the Expression

Simplify the expression: \( f'(x) = (2x^2 - 3) + (4x^2 + 8x) = 6x^2 + 8x - 3 \).

Key Concepts

Product Rule

The product rule is an essential tool in calculus differentiation. It provides a method to find the derivative of the product of two functions. The rule states that if you have two functions, let's name them \(u(x)\) and \(v(x)\), then the derivative of the product \(u(x) \cdot v(x)\) is given by: \[ (uv)' = u'v + uv' \]. Here, \( u'(x) \) and \( v'(x) \) denote the derivatives of \(u(x)\) and \(v(x)\) respectively. This means, to find the derivative of the product you:

• Differentiating the first function \(u(x)\) while keeping \(v(x)\) constant,
• Then, add the result to the product of the original function \(u(x)\) and the derivative of \(v(x)\).

Understanding when and how to apply the product rule is a key skill in calculus, as it helps address more complex expressions involving products of different types of functions.

Derivatives

Derivatives play a critical role in calculus, representing the rate of change or the slope of a function at any given point. Think of derivatives as a tool that tells you how quickly a quantity is changing. Calculating a derivative involves several rules and approaches, like:

• The power rule, used for polynomial functions,
• The product rule, for products of two functions,
• And others like the quotient rule or chain rule.

For a simple function such as \( u(x) = x + 2 \), differentiation is straightforward, giving \( u'(x) = 1 \). This demonstrates how with a linear function, you just take the coefficient of \(x\). For more complex functions like \( v(x) = 2x^2 - 3 \), you'd use the power rule and get \( v'(x) = 4x \), showing how the derivative of \(x^n\) is \(nx^{n-1}\). Understanding different techniques to compute derivatives allows you to tackle a variety of problems.

Polynomial Functions

Polynomial functions are expressions involving variables raised to non-negative integer powers. A typical polynomial looks like \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \). One of the main characteristics of polynomial functions is their smooth, continuous curves. Differentiation of polynomials is direct using the power rule. For example, consider \( v(x) = 2x^2 - 3 \), a simple quadratic polynomial. Differentiating each term separately, we apply the power rule to obtain \( v'(x) = 2 \cdot 2x^{2-1} = 4x \), while constants like \(-3\) have a derivative of zero. Polynomials are fundamental in calculus as virtually any function can be approximated by polynomials using Taylor Series. This makes learning to work with their derivatives a foundational skill in calculus differentiation.