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Find the derivative the following ways: a. Using the Product Rule (Exercises 7-10 ) or the Quotient Rule (Exercises 11-14 ). Simplify your result. b. By expanding the product first (Exercises \(7-10\) ) or by simplifying the quotient first (Exercises 11-14 ). Verify that your answer agrees with part ( \(a\) ). $$f(x)=(x-1)(3 x+4)$$

Short Answer

Expert verified
Question: Determine the derivative of the function \(f(x) = (x-1)(3x+4)\) using both the Product Rule and by expanding the product first. Verify that both methods yield the same result. Answer: The derivative of the function \(f(x) = (x-1)(3x+4)\) using both the Product Rule and by expanding the product first is \(\frac{d}{dx}f(x) = 6x + 1\).

Step by step solution

01

Identify the functions u(x) and v(x)

For our given function, \(f(x) = (x-1)(3x+4)\), we can identify the functions \(u(x)\) and \(v(x)\): $$u(x) = x-1$$ $$v(x) = 3x + 4$$
02

Find the derivatives of u(x) and v(x)

Next, we need to find the derivatives of \(u(x)\) and \(v(x)\): $$\frac{d}{dx}u(x) = \frac{d}{dx}(x-1) = 1$$ $$\frac{d}{dx}v(x) = \frac{d}{dx}(3x+4) = 3$$
03

Apply the Product Rule to find the derivative of f(x)

Now we can apply the Product Rule: $$\frac{d}{dx}f(x) = u(x)\frac{d}{dx}v(x) + v(x)\frac{d}{dx}u(x)$$ Plugging in all the expressions, we have: $$\frac{d}{dx}f(x) = (x-1)(3) + (3x + 4)(1)$$ Now we will simplify this expression: $$\frac{d}{dx}f(x) = 3x-3+3x+4 = 6x +1$$ b. By expanding the product first
04

Expand the product in f(x)

Before taking the derivative, let's first expand the given product \((x-1)(3x+4)\): $$f(x) = (x-1)(3x+4) = 3x^2 + 4x - 3x - 4 = 3x^2 + x - 4$$
05

Find the derivative of the expanded function

Now we can take the derivative of the expanded function: $$\frac{d}{dx}f(x) = \frac{d}{dx}(3x^2 + x - 4) = 3(2x)+1(1) = 6x+1$$ As we can see, both methods give us the same result for the derivative of the given function, which is: $$\frac{d}{dx}f(x) = 6x + 1$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The Product Rule is a fundamental concept in calculus, used to find the derivative of a product of two functions. If you have a function that is the product of two other functions, say \(u(x)\) and \(v(x)\), then the Product Rule provides a way to find the derivative of this product. Imagine you have \(f(x) = u(x)\cdot v(x)\). To get the derivative, you would use the formula:\[\frac{d}{dx}f(x) = u(x) \cdot \frac{d}{dx}v(x) + v(x) \cdot \frac{d}{dx}u(x)\]This formula combines the derivatives of \(u(x)\) and \(v(x)\) by multiplying each function by the derivative of the other and summing the results.
Consider the function from our exercise, \(f(x) = (x-1)(3x+4)\). Applying the Product Rule:
  • Identify \(u(x) = x - 1\) and \(v(x) = 3x + 4\).
  • Find their derivatives: \(\frac{d}{dx}u(x) = 1\) and \(\frac{d}{dx}v(x) = 3\).
  • Substitute into the Product Rule formula: \(u(x)\,d(v) + v(x)\,d(u) = (x-1) \cdot 3 + (3x+4) \cdot 1\).
  • Finally, simplify to get \(6x + 1\).
By following these steps, you can efficiently find the derivative of any product of two functions.
Quotient Rule
The Quotient Rule is a method for differentiating functions that are in the form of one function divided by another, symbolized as \( \frac{u(x)}{v(x)} \). It is crucial when simplifying such functions before differentiation isn't enough or when a derivative is needed quickly.
The formula for the Quotient Rule is:\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x)\cdot \frac{d}{dx}u(x) - u(x)\cdot \frac{d}{dx}v(x)}{(v(x))^2}\]This formula might look intimidating, but it's just carefully applying the derivatives of the numerator and the denominator:
  • Differentiate the numerator \(u(x)\).
  • Differentiate the denominator \(v(x)\).
  • Multiply the derivative of the denominator by the numerator function \(u(x)\), then subtract from it the product of the derivative of the numerator by the denominator.
  • Finally, divide the entire expression by the square of the denominator \(v(x)^2\).
This process provides a systematic way to find the derivatives of functions that are quots, making it a powerful tool for calculus students.
Function Expansion
Function expansion is a technique used to rewrite a function in a simpler form that makes it easier to take its derivative, integrate, or simplify further.
In the exercise, we were asked to expand the function \((x-1)(3x+4)\). This involves distributive property: multiplying each term in one polynomial by each term of the other polynomial.
Steps for expanding \((x-1)(3x+4)\):
  • First, distribute \((x - 1)\) across every term in \((3x + 4)\).
  • Calculate \(3x \cdot x\) and \( 3x \cdot (-1)\) to get \(3x^2 - 3x\).
  • Next, do \(4 \cdot x\) and \(4 \cdot (-1)\) to get \(4x - 4\).
  • Combine all terms to form the expanded function \(3x^2 + x - 4\).
Once expanded, finding the derivative becomes straightforward, using basic rules of differentiation on each term individually.

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