Chapter 3: Problem 31
Find \(D_{x} y\) using the rules of this section. $$ y=\left(5 x^{2}-7\right)\left(3 x^{2}-2 x+1\right) $$
Short Answer
Expert verified
The derivative is \( D_x y = 60x^3 - 30x^2 - 32x + 14 \).
Step by step solution
01
Identify the Product Rule
The function you are given, \( y = (5x^2 - 7)(3x^2 - 2x + 1) \), is a product of two functions: \( u = 5x^2 - 7 \) and \( v = 3x^2 - 2x + 1 \). The Product Rule states that the derivative of a product \( uv \) is \( u'v + uv' \).
02
Differentiate Each Function
Differentiate \( u = 5x^2 - 7 \) and \( v = 3x^2 - 2x + 1 \) separately. First, the derivative of \( u \) with respect to \( x \) is \( u' = 10x \). Second, differentiate \( v \) to obtain \( v' = 6x - 2 \).
03
Apply the Product Rule
Now, substitute \( u' = 10x \), \( u = 5x^2 - 7 \), \( v' = 6x - 2 \), and \( v = 3x^2 - 2x + 1 \) into the product rule formula: \( D_x y = (10x)(3x^2 - 2x + 1) + (5x^2 - 7)(6x - 2) \).
04
Simplify Each Term
Expand each term: - First term: \( (10x)(3x^2 - 2x + 1) = 30x^3 - 20x^2 + 10x \).- Second term: \( (5x^2 - 7)(6x - 2) = 30x^3 - 10x^2 - 42x + 14 \).
05
Combine and Simplify
Add the two expanded terms from Step 4: \( 30x^3 - 20x^2 + 10x \) and \( 30x^3 - 10x^2 - 42x + 14 \). Combine like terms to simplify:\( 30x^3 + 30x^3 - 20x^2 - 10x^2 + 10x - 42x + 14 = 60x^3 - 30x^2 - 32x + 14 \).
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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 an essential tool in calculus for finding the derivative of functions that are the product of two or more functions. It's like a secret recipe that helps you differentiate products without expanding them completely first. When you have a function that looks like this:
- \( y = u(x) \cdot v(x) \)
- \( D_x y = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
Derivative
A derivative represents the rate at which a function is changing at any given point and is one of the central ideas of calculus. If you think of graphs, the derivative is the slope of the tangent line to the curve at a particular point. This concept lets us understand how a function behaves and predicts things like speed or growth.In mathematical terms, if you have a function \( f(x) \), its derivative is denoted as \( f'(x) \) or \( \frac{df}{dx} \). The process of finding the derivative is called differentiation. Differentiation provides valuable insights, such as:
- Finding maximum or minimum values of functions
- Understanding how points are connected and how steep they are
- Calculating acceleration or any other rate of change
Polynomial Functions
Polynomial functions are mathematical expressions involving variables raised to non-negative integer powers. They take a form like \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where \( a_n, a_{n-1}, \ldots, a_0 \) are constants and \( n \) is a non-negative integer.These functions are straightforward to work with in calculus because they obey familiar rules, making them perfect candidates for differentiation. The degree of the polynomial (the highest power of \( x \)) informs us of the behavior and shape of their graphs:
- Linear (straight line), when the highest degree is 1
- Quadratic (parabolic curve), when the highest degree is 2
- Cubic (S-shaped curve), when the highest degree is 3, and so on
Differentiation Techniques
Differentiation techniques are the methods used to find derivatives of functions. Some basic rules guide this process, and mastering these can make the journey through calculus smoother. Major techniques include:
- Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- Constant Rule: The derivative of a constant is 0.
- Sum and Difference Rules: The derivative of the sum/difference of functions is the sum/difference of their derivatives.
- Product Rule: As we've already discussed, it's for differentiating products of functions.
- Quotient Rule: For quotients, where you have: \( \frac{u}{v} \), use: \( \frac{v \cdot u' - u \cdot v'}{v^2} \).