Problem 41 - Calculus an Applied Approach

Question

Find \(f^{\prime}(x)\) $$ f(x)=(x+4)\left(2 x^{2}-1\right) $$

Step-by-Step Solution

Answer

\[f'(x) = 6x^{2} + 16x - 1\]

Step 1: Identify the functions u(x) and v(x)

In this question, it can be observed that two functions are being multiplied. Therefore, we identify two functions:\n\(u(x) = x + 4\) and \(v(x) = 2x^{2} - 1\)

Step 2: Calculate the derivatives of the functions

For the function \(u(x) = x + 4\), the derivative is \(u'(x) = 1\), simply the derivative of x.\nFor the function \(v(x) = 2x^{2} - 1\), the derivative is \(v'(x) = 4x\), applying the power rule for differentiation.

Step 3: Apply the product rule

Now we apply the product rule for differentiation: \(f'(x) = u'(x).v(x) + u(x).v'(x)\). Substituting the derivatives and original functions we derived earlier into the formula, we end up with \(f'(x) = 1*(2x^{2} - 1) + (x + 4)*4x\)

Step 4: Simplification

The last step is to simplify the equation. Distribute all the terms properly to achieve the final solution: \(f'(x) = 2x^{2} - 1 + 4x^{2} + 16x = 6x^{2} + 16x - 1\)

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to differentiate products of two functions. When you're given two functions, say \( u(x) \) and \( v(x) \), and you need the derivative of their product \( u(x)v(x) \), the product rule comes in handy. It states that the derivative of this product is:

• \( (uv)' = u'v + uv' \)

This means you differentiate one function at a time while multiplying it by the other function, then swap roles and add the results. This rule is crucial for handling expressions where simple derivative rules don't apply directly, such as when more than one function is involved. Using the product rule helps to ensure you're accounting for how both functions in a product individually affect the derivative.

Differentiation

Differentiation is the process through which we find a derivative, which measures how a function changes as its input changes. In simple terms, the derivative of a function gives us the rate of change or the slope of the function at any given point. This operation is essential for a wide array of applications, from physics to economics.
Differentiation involves applying certain rules depending on the kind of functions you're working with. The key rules include:

• The sum/difference rule, where you differentiate terms individually.
• The product rule for products of functions.
• The quotient rule for divisions of functions.
• The chain rule for compositions of functions.

Understanding these rules allows us to differentiate complex expressions step-by-step, simplifying as we proceed to get to our answer. Each rule of differentiation provides insights on how to deal with different forms and combinations of functions.

Power Rule

The power rule is a straightforward yet powerful tool in calculus for finding derivatives. Specifically useful when dealing with polynomial functions, the power rule makes differentiation a breeze. The rule states:

• If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).

This means you bring down the exponent as a coefficient and reduce the exponent by one. For example, given a function like \( v(x) = 2x^2 - 1 \), using the power rule gives us \( v'(x) = 4x \), simplifying the process significantly.
The power rule is heavily utilized because many functions in calculus are polynomials or contain polynomial parts. It's important to remember that the power rule only applies to terms where the variable is raised to a constant power, making it one of the most frequently used and fundamental rules in differentiation.