Problem 9
Question
Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(y=3 x-2 \cdot 4^{x}\)
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = 3 - 2 \cdot 4^x \ln(4) \).
1Step 1: Identify the components of the function
The given function is \( y = 3x - 2 \cdot 4^x \). This function consists of two parts: a linear part \( 3x \) and an exponential part \( -2 \cdot 4^x \). Our goal is to differentiate this entire function with respect to \( x \).
2Step 2: Differentiate the linear term
The linear term is \( 3x \). The derivative of \( 3x \) with respect to \( x \) is simply the constant multiplying \( x \). Therefore, \( \frac{d}{dx}(3x) = 3 \).
3Step 3: Differentiate the exponential term using chain rule
The exponential term is \( -2 \cdot 4^x \). Using the chain rule, the derivative of \( 4^x \) is \( 4^x \ln(4) \). Therefore, the derivative of \( -2 \cdot 4^x \) is \( -2 \times 4^x \ln(4) \).
4Step 4: Combine the derivatives
Combine the derivatives from Steps 2 and 3 to get the derivative of the entire function:\[ \frac{dy}{dx} = 3 - 2 \cdot 4^x \ln(4) \].
Key Concepts
Linear FunctionExponential FunctionChain Rule
Linear Function
Linear functions are the simplest types of functions you will encounter in mathematics. They have the form \( y = mx + c \), where \( m \) is the slope of the line and \( c \) is the y-intercept. The slope \( m \) tells us how steep the line is and in which direction it moves. It is a constant value, and this makes linear functions easy to differentiate. Since differentiating means finding out how a function changes, when applied to linear functions, the derivative is simply the constant slope \( m \).
In the given function \( y = 3x - 2 \cdot 4^x \), the linear part is \( 3x \). The derivative of \( 3x \) is \( 3 \), because the derivative of \( x \) with respect to \( x \) is 1, and multiplying that by 3 gives us 3. Thus, linear functions have very straightforward derivatives.
In the given function \( y = 3x - 2 \cdot 4^x \), the linear part is \( 3x \). The derivative of \( 3x \) is \( 3 \), because the derivative of \( x \) with respect to \( x \) is 1, and multiplying that by 3 gives us 3. Thus, linear functions have very straightforward derivatives.
Exponential Function
Exponential functions involve terms where the variable is an exponent, typically written as \( a^x \), where \( a \) is a positive constant. These functions have unique properties, such as rapid growth rates and the fact that their derivatives are proportional to their values.
In the exercise function \( y = 3x - 2 \cdot 4^x \), the exponential part is \( -2 \cdot 4^x \). To differentiate this part, you need to apply knowledge of the exponential derivative: the derivative of \( a^x \) is \( a^x \ln(a) \). Here, for \( 4^x \), it becomes \( 4^x \ln(4) \).
Thus, multiplying by \( -2 \) gives the derivative \( -2 \times 4^x \ln(4) \). Exponential derivatives show how quickly the function increases or decreases as \( x \) changes.
In the exercise function \( y = 3x - 2 \cdot 4^x \), the exponential part is \( -2 \cdot 4^x \). To differentiate this part, you need to apply knowledge of the exponential derivative: the derivative of \( a^x \) is \( a^x \ln(a) \). Here, for \( 4^x \), it becomes \( 4^x \ln(4) \).
Thus, multiplying by \( -2 \) gives the derivative \( -2 \times 4^x \ln(4) \). Exponential derivatives show how quickly the function increases or decreases as \( x \) changes.
Chain Rule
The chain rule is a fundamental technique used to differentiate composite functions. A composite function is essentially a function within a function. The chain rule helps us differentiate these by focusing on solving each part step-by-step and then multiplying the derivatives.
The rule can be stated as: if you have a function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our problem, the exponential term \( 4^x \) needed the chain rule for differentiation. By recognizing \( 4^x \) as an exponential function \( e^{x \ln(4)} \), you differentiate the outer function while holding the inside function \( x \ln(4) \) and multiply by the derivative of the inside, which is \( \ln(4) \). So, the application of the chain rule allows the derivative to be \( 4^x \ln(4) \).
This rule is invaluable in calculus as it simplifies the process of dealing with nested functions, enabling us to tackle more complex problems.
The rule can be stated as: if you have a function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our problem, the exponential term \( 4^x \) needed the chain rule for differentiation. By recognizing \( 4^x \) as an exponential function \( e^{x \ln(4)} \), you differentiate the outer function while holding the inside function \( x \ln(4) \) and multiply by the derivative of the inside, which is \( \ln(4) \). So, the application of the chain rule allows the derivative to be \( 4^x \ln(4) \).
This rule is invaluable in calculus as it simplifies the process of dealing with nested functions, enabling us to tackle more complex problems.
Other exercises in this chapter
Problem 9
Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ W=4 \cos \left(t^{2}\right) $$
View solution Problem 9
Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ z=(3 t+1)(5 t+2) $$
View solution Problem 9
Find the derivative. Assume \(a, b, c, k\) are constants. $$f(x)=\frac{1}{x^{4}}$$
View solution Problem 10
Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ y=2 \cos (5 t) $$
View solution