Problem 5 - Calculus for Biology and Medicine

Question

Use the product rule to find the derivative with respect to the independent variable. $$ f(x)=\left(\frac{1}{2} x^{2}-1\right)\left(2 x+3 x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 4x^3 + 3x^2 - 6x - 2 \).

Step 1: Identify Functions

Recognize the two functions we need to differentiate using the product rule. Let \( u(x) = \frac{1}{2} x^2 - 1 \) and \( v(x) = 2x + 3x^2 \). We need to compute the derivative of \( f(x) = u(x) \cdot v(x) \).

Step 2: Differentiate each Function

Find the derivatives of \( u(x) \) and \( v(x) \):\-\( u'(x) = \frac{d}{dx}\left(\frac{1}{2}x^2 - 1\right) = x \).\( v'(x) = \frac{d}{dx}(2x + 3x^2) = 2 + 6x \).

Step 3: Apply the Product Rule

According to the product rule, the derivative of \( f(x) = u(x) \cdot v(x) \) is \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Step 4: Substitute the Derivatives

Substitute the derivatives \( u'(x) \), \( v'(x) \) as well as \( u(x) \) and \( v(x) \) into the product rule formula:\[f'(x) = x(2x + 3x^2) + \left(\frac{1}{2}x^2 - 1\right)(2 + 6x)\].

Step 5: Expand Terms

Expand both terms in the expression:- First term: \( x(2x + 3x^2) = 2x^2 + 3x^3 \).- Second term: \( \left(\frac{1}{2}x^2 - 1\right)(2 + 6x) = \frac{1}{2}x^2(2) + \frac{1}{2}x^2(6x) - 1(2) - 1(6x) = x^2 + 3x^3 - 2 - 6x \).

Step 6: Combine Like Terms

Combine the expanded terms:\[f'(x) = (2x^2 + 3x^3) + (x^2 + 3x^3 - 2 - 6x) = 4x^3 + 3x^2 - 6x - 2 \]

Step 7: Final Simplification

The final expression is already simple enough after combining like terms. The derivative of the function is:\[f'(x) = 4x^3 + 3x^2 - 6x - 2.\]

Key Concepts

Differentiation

Differentiation is a key concept in calculus. It involves finding the rate at which a function is changing at any given point. This process is known as calculating the function's derivative.
In simple terms, differentiation is about finding the slope of the tangent line to the function at any point. Imagine roads twisting and turning in a hilly area; differentiation helps determine how steep or flat the road is at any particular location.
By knowing how the function behaves, we can make predictions and understand its properties better.
Steps for differentiation usually involve breaking down a complex problem into simpler parts, much like tackling a jigsaw puzzle piece by piece. Techniques like the product rule, quotient rule, and chain rule are often employed, especially when dealing with polynomials and algebraic expressions.
Mastering differentiation opens up a wealth of possibilities in math and science, from engineering to physical sciences, where understanding change and slopes is vital.

Derivative

The derivative is the core tool used in differentiation. It represents the instantaneous rate of change of a function with respect to one of its variables.
Think of it as the speedometer in a car, which tells us the speed at any given moment, not just the start or finish speed.
Some important aspects of derivatives include:

• The derivative of a constant is zero. This is because constants do not change.
• The power rule: If you have a term like \(x^n\), its derivative is \(nx^{n-1}\).
• For more complex functions, we often use the product rule or the chain rule, depending on the interaction between the components.

In our problem, the product rule helps find the derivative of a function that is the product of two simpler functions. By understanding how each part changes, we can deduce the rate of change for the entire expression. This is a vital skill for tackling real-world problems modeled by complex functions.

Algebraic Expressions

Algebraic expressions are at the heart of many mathematical problems, commonly appearing in calculus in the form of polynomials, fractions, and equations.
Understanding how to manipulate these expressions is essential, as they often need to be simplified or transformed to solve calculus problems.
Some key aspects to consider when dealing with algebraic expressions include:

• Simplifying by combining like terms. This helps in reducing expressions to their simplest form.
• Expanding expressions, especially when using rules like the product rule, which necessitates dealing with products of two or more terms.
• Factoring expressions, which can simplify both integration and differentiation tasks.

In our exercise, we used algebraic manipulation to break down the derivative after applying the product rule. This simplification made it easier to understand and work with, allowing for effective problem-solving techniques.