Problem 5 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find the derivative of each of the following functions. $$f(x)=(3 x-2)^{5}\left(x^{2}-1\right)$$

Step-by-Step Solution

Answer

Answer: The derivative of the given function is \(f'(x) = 15(3x-2)^4(x^2-1) + (3x-2)^5(2x)\).

Step 1: Identify the two functions

We have a product of two functions: $$u(x) = (3x-2)^5$$ $$v(x) = x^2-1$$

Step 2: Find the derivatives of both functions

Following the chain rule: $$u'(x) = 5(3x-2)^4 \cdot 3$$ Simplifying gives: $$u'(x) = 15(3x-2)^4$$ Now, applying power rule for v(x): $$v'(x) = 2x$$

Step 3: Apply the product rule

Now use the product rule: \((uv)' = u'v + uv'\) $$f'(x) = 15(3x-2)^4(x^2-1) + (3x-2)^5(2x)$$ The derivative of the given function is $$f'(x) = 15(3x-2)^4(x^2-1) + (3x-2)^5(2x)$$

Key Concepts

Understanding the Chain Rule

The chain rule is an essential technique when you're working with composite functions, meaning a function inside another function. Imagine peeling an onion, layer by layer; that's similar to how the chain rule works, by handling each 'layer' of the function.

When applying the chain rule, you differentiate the outer function while keeping the inner function intact. Then, follow by multiplying with the derivative of the inner function. You can picture this like, "Take the shell off, then peel the inside."

• First, identify the outer and inner functions.
• Differentiating the outer function: Treat the inner function as a constant.
• Multiply the result by the derivative of the inner function.

In our exercise, with \(u(x) = (3x-2)^5\), we use the chain rule. By differentiating the outer exponential function, we keep \(3x-2\) intact, resulting in \((3x-2)^4\). Next, differentiate the inner \(3x-2\), resulting in a factor of 3. Finally, multiply these results to obtain \(u'(x) = 15(3x-2)^4\). This systematic approach helps break down complex expressions into manageable parts.

The Power of the Power Rule

The power rule is one of the simplest yet most powerful tools for finding derivatives, especially when dealing with expressions like \(x^n\). It allows you to differentiate powers of x quickly and efficiently. Just remember, "bring down the power, reduce it by one."

• For \(x^n\), the derivative is \(nx^{n-1}\).
• This means you take the exponent in front and lower the exponent on the variable by 1.

In our problem, we apply the power rule to derive \(v(x) = x^2 - 1\). Taking the derivative of \(x^2\), bring down the 2 and subtract 1 from the exponent, leading us to \(v'(x) = 2x\). The constant \(-1\) vanishes since it does not change as x changes. Remember, the power rule is about adjusting the exponent and making calculation straightforward.

Applying the Product Rule

The product rule helps us differentiate functions that are multiplied together. Think of it this way: when you have a product, you may focus on one part while holding the other constant, then switch. It’s like being fair to both functions in play.

The product rule formula is \[ (uv)' = u'v + uv' \] where \(u\) and \(v\) are functions of \(x\). To use this, differentiate one function while keeping the other constant, then reverse the roles:

• Differentiate \(u(x)\): Find \(u'(x)\)
• Multiply \(u'(x)\) by \(v(x)\)
• Differentiate \(v(x)\): Get \(v'(x)\)
• Multiply \(v'(x)\) by the original \(u(x)\)
• Add the results together.

In our task, we used \(u(x) = (3x-2)^5\) and \(v(x) = x^2 - 1\). We calculated \(u'(x) = 15(3x-2)^4\) and \(v'(x) = 2x\). Using these, the product rule gives \[ f'(x) = 15(3x-2)^4(x^2-1) + (3x-2)^5(2x) \]. This technique ensures every changing part of the function is considered.