Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\left(3 x^{3}+4 x\right)(x-5)(x+1) $$

Short answer

Applying the product rule twice on this function by pairing \((3x^3 + 4x)\) with \((x-5)\) first, and then their product with \((x+1)\), and using the power rule to find derivatives of the individual terms, one eventually arrives at the final derivative.

Step-by-step solution

Identify Relevant Differentiation Rules

The function is a product of three functions and thus requires the Product Rule for differentiation. However, as there are three functions multiplied together, it will need to be applied twice. Additionally, as the individual terms within each function involve powers of \(x\), the Power Rule will also be utilized.

Apply the First Product Rule

Taking the derivative of \((3x^3 + 4x)(x-5)\) and \((x+1)\) separately, first apply the product rule on the first pair of functions. The Product Rule states that the derivative of two functions multiplied together is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. It's written as, \(g(x) * h'(x) + g'(x) * h(x)\). Thus, \((3x^3 + 4x)(x-5)\) using the product rule becomes \(3x^3(x-5)' + (3x^3+4x)'(x-5)\).

Apply the Power Rule

Then apply the Power Rule on the two parts in the above expression. The Power Rule states that the derivative of \(x^n\) is \(n*x^{(n-1)}\). Applying this gives, \((3*3*x^{(3-1)}+4)(x-5) + 4x'*(x-5)\). Simplifying this results in \(9x^2*(x-5) + 4*(x-5)\).

Apply the Product Rule Again

Now one has to apply the product rule again. Here, the product rule will be applied to the result from the previous steps and \((x+1)\). This becomes, \((9x^2*(x-5) + 4*(x-5))*(x+1)' + (9x^2*(x-5) + 4*(x-5))'*(x+1)\).

Simplify the Resulting Expression

Once again apply the Power Rule and simplify the above expression to its simplest form which will be the final derivative of the function \(f(x)\).

Key concepts

Understanding the Product Rule

When you have to find the derivative of a function that is a product of two or more functions, the Product Rule becomes your best friend. Simply put, the Product Rule helps us differentiate functions that are multiplied together. If you have two functions, say \( g(x) \) and \( h(x) \), then the derivative of their product is:

• \( (g \cdot h)' = g \cdot h' + g' \cdot h \)

This means you take the derivative of the first function and multiply it by the second function, then add the first function times the derivative of the second.
In the exercise, we had three functions multiplied together. This means applying the Product Rule more than once, carefully following this pattern. It's crucial to handle each part one at a time to ensure accuracy.

Mastering the Power Rule

The Power Rule is one of the simplest yet powerful tools we have in calculus for differentiation. It directly applies when you have a term with \( x \) raised to a power. The rule states:

• \( \frac{d}{dx} [x^n] = n \cdot x^{n-1} \)

This means the exponent comes down as a coefficient, and then we reduce the exponent by one.
In the exercise, we've applied the Power Rule to the terms \( 3x^3 \) and \( 4x \). For \( 3x^3 \) the derivative becomes \( 9x^2 \) because \( 3 \cdot 3 = 9 \) and the new power of \( x \) becomes 2.
Remember, whenever you encounter terms like these, the Power Rule is your go-to method.

What is a Derivative Anyway?

At the heart of calculus is the concept of the derivative, which represents the rate of change of a function with respect to one of its variables. If you think of a graph, the derivative gives you the slope of the tangent line at any given point of a function.
In practical terms, it helps us understand how a function is changing. For instance, if you're looking at a car's position over time, the derivative tells you the speed of the car at any moment.
In our exercise, applying the Product Rule and the Power Rule allowed us to find the derivative of the complex function, showing us how it behaves dynamically with respect to \( x \). Understanding derivatives is crucial for deeper insights into how functions operate and evolve.

The Big Picture: Calculus

Calculus is the mathematical study of change. With tools like derivatives, calculus lets us delve into how quantities vary together. Whether it's physics, engineering, or economics, calculus helps model real-world processes.
Differentiation, which involves finding derivatives, is just one aspect. We also have integration, which provides ways to calculate areas and other quantities under curves.
In essence, calculus offers powerful methods for analyzing systems and solving complex problems. It equips you with the tools to not only understand motion and change but to predict and optimize as well.