Problem 40
Question
Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\left(3 x^{3}+4 x\right)(x-5)(x+1) $$
Step-by-Step Solution
Verified 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.
1Step 1: 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.
2Step 2: 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 \(3x^3(x-5)' + (3x^3+4x)'(x-5)\).
3Step 3: 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 \(9x^2*(x-5) + 4*(x-5)\).
4Step 4: 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, \((9x^2*(x-5) + 4*(x-5))*(x+1)' + (9x^2*(x-5) + 4*(x-5))'*(x+1)\).
5Step 5: 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 RuleMastering the Power RuleWhat is a Derivative Anyway?The Big Picture: Calculus
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:
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.
- \( (g \cdot h)' = g \cdot h' + g' \cdot h \)
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:
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.
- \( \frac{d}{dx} [x^n] = n \cdot x^{n-1} \)
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.
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.
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.
Other exercises in this chapter
Problem 39
Find the limit. $$ \lim _{x \rightarrow 1} \frac{\frac{1}{x+4}-\frac{1}{4}}{x} $$
View solution Problem 40
Use the General Power Rule to find the derivative of the function. $$ f(x)=(4-3 x)^{-5 / 2} $$
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The cost \(C\) of producing \(x\) units is modeled by \(C=v(x)+k\), where \(v\) represents the variable cost and \(k\) represents the fixed cost. Show that the
View solution Problem 40
find \(f^{\prime}(x)\). $$ f(x)=\left(x^{2}+2 x\right)(x+1) $$
View solution