Chapter 23: Problem 25
Evaluate the derivatives of the given functions for the given values of \(x\). Use the product rule. Check your results using the derivative evaluation feature of a calculator. $$y=(3 x-1)(4-7 x), x=3$$
Short Answer
Expert verified
The derivative of the given function at \( x = 3 \) is \( -107 \).
Step by step solution
01
Identify the Functions
The given function is a product of two functions: \( u(x) = 3x - 1 \) and \( v(x) = 4 - 7x \). We will use the product rule, which states that the derivative of a product of two functions \( y = u(x)v(x) \) is \( y' = u'(x)v(x) + u(x)v'(x) \).
02
Differentiate the Functions
First, find the derivatives of \( u(x) \) and \( v(x) \). The derivative of \( u(x) = 3x - 1 \) is \( u'(x) = 3 \). The derivative of \( v(x) = 4 - 7x \) is \( v'(x) = -7 \).
03
Apply the Product Rule
Using the product rule, substitute the derivatives and original functions: \( y' = u'(x)v(x) + u(x)v'(x) \). Thus, \( y' = 3(4 - 7x) + (3x-1)(-7) \).
04
Simplify the Expression
Simplify the expression to find \( y' \). First, expand each term: \( 3(4 - 7x) = 12 - 21x \) and \( (3x-1)(-7) = -21x + 7 \). Combine these results: \( y' = (12 - 21x) + (-21x + 7) = 19 - 42x \).
05
Evaluate the Derivative at x=3
Substitute \( x = 3 \) into the derivative: \( y' = 19 - 42(3) = 19 - 126 = -107 \). The derivative at \( x = 3 \) is \( -107 \).
06
Verify with a Calculator
Use a calculator's derivative evaluation feature to verify this result. Input the function \( y = (3x-1)(4-7x) \) and evaluate the derivative at \( x = 3 \). You should obtain \( -107 \), confirming our manual calculation.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
In calculus, a derivative represents the rate of change of a function with respect to a variable. Essentially, derivatives tell us how a function's output values change as its input values change. When working with derivatives, you're often trying to find how quickly something changes. For instance, in motion, a derivative of position with respect to time gives us velocity.
Understanding derivatives is crucial because they form the foundation for many aspects of calculus and its applications:
- Finding slopes of tangent lines at any point on a curve
- Determining maxima and minima of functions
- Analyzing dynamic changes in various scientific fields
Product of Functions and the Product Rule
The product rule is a differentiation technique used when dealing with the product of two separate functions. For any functions \( u(x) \) and \( v(x) \), the product \( y = u(x)v(x) \) can be differentiated using the formula:\[ y' = u'(x)v(x) + u(x)v'(x) \]This formula is essential when you cannot simply multiply the functions and differentiate the result directly, especially in more complex problems where simplification isn't straightforward.Here's a breakdown of how to use the product rule:
- Identify the two functions being multiplied together.
- Calculate the derivative of each individual function separately.
- Substitute these derivatives into the product rule formula.
- Simplify and solve for the desired derivative of the entire product.
Differentiation Techniques in Calculus
Differentiation techniques are methods used to calculate the derivative of functions. These can include a wide array of rules beyond the product rule, catering to different types and combinations of functions. Here are some important differentiation techniques:
- Constant Rule: The derivative of a constant is always zero.
- Power Rule: For \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
- Chain Rule: Used to differentiate composite functions or functions within other functions. For \( f(g(x)) \), the derivative is \( f'(g(x))g'(x) \).
- Quotient Rule: Handles the division of two functions. If \( y = \frac{u(x)}{v(x)} \), then \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).