Chapter 2: Problem 37
Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=(2 x+1)^{3}(2 x-1)^{4} $$
Short Answer
Expert verified
The derivative is \((2x+1)^2(2x-1)^3 (28x+2)\).
Step by step solution
01
Identify the Structure
To find the derivative of the function \( f(x)=(2x+1)^3(2x-1)^4 \), we first note that this is a product of two functions, \( u(x)=(2x+1)^3 \) and \( v(x)=(2x-1)^4 \). Therefore, we'll need to use the product rule in conjunction with the generalized power rule.
02
Apply the Product Rule
The product rule states that if \( f(x) = u(x)v(x) \), then the derivative \( f'(x) = u'(x)v(x) + u(x)v'(x) \). So, we'll compute \( u'(x) \) and \( v'(x) \) separately.
03
Differentiate \( u(x) \) Using the Generalized Power Rule
For \( u(x) = (2x+1)^3 \), use the generalized power rule: \( \frac{d}{dx}[(g(x))^n] = n(g(x))^{n-1}g'(x) \). Here, \( n = 3 \) and \( g(x) = 2x+1 \). Thus, \( u'(x) = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2 \), where \( g'(x) = 2 \).
04
Differentiate \( v(x) \) Using the Generalized Power Rule
For \( v(x) = (2x-1)^4 \), using the same rule, \( n = 4 \) and \( g(x) = 2x-1 \). So, \( v'(x) = 4(2x-1)^3 \cdot 2 = 8(2x-1)^3 \), where \( g'(x) = 2 \).
05
Substitute into the Product Rule Formula
Substitute \( u'(x) \), \( v'(x) \), \( u(x) \), and \( v(x) \) back into the product rule formula: \[ f'(x) = 6(2x+1)^2(2x-1)^4 + (2x+1)^3 \times 8(2x-1)^3 \]
06
Simplify the Expression
Both terms of \( f'(x) \) share common factors. To simplify, factor out the greatest common factor: \( (2x+1)^2(2x-1)^3 \). Thus, \[ f'(x) = (2x+1)^2(2x-1)^3 \left[6(2x-1) + 8(2x+1)\right] \]. Simplify the expression inside the brackets: \[ 6(2x-1) + 8(2x+1) = 12x - 6 + 16x + 8 = 28x + 2 \]. So, the simplified derivative is \[ f'(x) = (2x+1)^2(2x-1)^3 (28x+2) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
The concept of a derivative is foundational in calculus. A derivative measures how a function changes as its input changes. This can be thought of as the slope of the function's graph at any given point. For instance, if you have a function representing the distance an object travels over time, the derivative would give you the object's speed.Mathematically, the derivative is found using a process called differentiation. The derivative of a function is denoted by symbols such as \( f'(x) \) or \( \frac{df}{dx} \). It tells us the rate of change of the function with respect to its variable, typically \( x \).
- If the derivative is positive, the function is increasing.
- If the derivative is negative, the function is decreasing.
- If the derivative is zero, the function may have a local maximum, minimum, or a point of inflection.
Exploring the Product Rule
When a function is composed of two or more multiplied functions, like \( f(x) = u(x) \cdot v(x) \), we need to use the product rule to find its derivative. The product rule is a key tool in calculus that allows us to differentiate products of functions.The product rule states:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
- \( u'(x) \) is the derivative of the first function, \( u(x) \).
- \( v'(x) \) is the derivative of the second function, \( v(x) \).
Mastering the Differentiation Process
Differentiation is the process of finding the derivative of a function. This involves a series of steps and the application of various rules to simplify and solve.For power functions, the Generalized Power Rule is particularly helpful. This rule states that the derivative of a function \( (g(x))^n \) is:\[ \frac{d}{dx}[(g(x))^n] = n(g(x))^{n-1}g'(x) \]
- \( n \) is the exponent.
- \( g(x) \) is the inner function.
- \( g'(x) \) is the derivative of the inner function.