Problem 16
Question
Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ f(x)=2 x \sin (3 x) $$
Step-by-Step Solution
Verified Answer
The derivative is \( f'(x) = 2\sin(3x) + 6x\cos(3x) \).
1Step 1: Identify the Rule
To differentiate the function \( f(x) = 2x \sin(3x) \), notice that it is the product of two functions, \( u = 2x \) and \( v = \sin(3x) \). Therefore, we will use the product rule for differentiation: \( (uv)' = u'v + uv' \).
2Step 2: Differentiate First Function
Differentiate the first function \( u = 2x \). The derivative is \( u' = 2 \).
3Step 3: Differentiate Second Function
Differentiate the second function \( v = \sin(3x) \). Using the chain rule, the derivative is \( v' = 3 \cos(3x) \).
4Step 4: Apply the Product Rule
Substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule formula: \( f'(x) = u'v + uv' = 2\sin(3x) + 2x \cdot 3\cos(3x) \).
5Step 5: Simplify the Expression
Simplify the expression obtained in the previous step: \( f'(x) = 2\sin(3x) + 6x\cos(3x) \).
Key Concepts
Product RuleChain RuleDerivative
Product Rule
In the realm of calculus, the product rule is indispensable when differentiating the product of two functions. It provides a systematic way to find the derivative, ensuring you don't miss any components.
The product rule states that for two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( uv \), denoted \( (uv)' \), is given by:
\[(uv)' = u'v + uv'\]
Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. This means you differentiate \( u \), multiply it by \( v \), and add it to the product of \( u \) and the derivative of \( v \).
Take the example function \( f(x) = 2x \sin(3x) \). Here, \( u = 2x \) and \( v = \sin(3x) \). Differentiating \( u \) gives \( u' = 2 \). Next, use the chain rule to differentiate \( v \) as described shortly. Finally, apply these derivatives back into the product rule formula to find the derivative of the entire function.
The product rule states that for two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( uv \), denoted \( (uv)' \), is given by:
\[(uv)' = u'v + uv'\]
Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. This means you differentiate \( u \), multiply it by \( v \), and add it to the product of \( u \) and the derivative of \( v \).
Take the example function \( f(x) = 2x \sin(3x) \). Here, \( u = 2x \) and \( v = \sin(3x) \). Differentiating \( u \) gives \( u' = 2 \). Next, use the chain rule to differentiate \( v \) as described shortly. Finally, apply these derivatives back into the product rule formula to find the derivative of the entire function.
Chain Rule
The chain rule is a powerful tool in calculus used when dealing with composite functions. It allows us to differentiate a function based on the derivative of its inner and outer components.
For a function \( y = g(f(x)) \), the chain rule states:
\[y' = g'(f(x)) \, f'(x)\]
This means you first differentiate the outer function \( g \) while keeping the inner function unchanged, and then multiply it by the derivative of the inner function \( f \).
In our step-by-step exercise, \( v = \sin(3x) \) is differentiated using the chain rule. Here, the outer function is \( \sin(x) \) and the inner function is \( 3x \). The derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( 3x \) is \( 3 \). Combine these using the chain rule to find \( v' = 3 \cos(3x) \). This component is crucial when applying the product rule to the whole function \( f(x) \).
For a function \( y = g(f(x)) \), the chain rule states:
\[y' = g'(f(x)) \, f'(x)\]
This means you first differentiate the outer function \( g \) while keeping the inner function unchanged, and then multiply it by the derivative of the inner function \( f \).
In our step-by-step exercise, \( v = \sin(3x) \) is differentiated using the chain rule. Here, the outer function is \( \sin(x) \) and the inner function is \( 3x \). The derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( 3x \) is \( 3 \). Combine these using the chain rule to find \( v' = 3 \cos(3x) \). This component is crucial when applying the product rule to the whole function \( f(x) \).
Derivative
A derivative represents the rate at which a function changes at any given point and is fundamental to calculus. It shows how sensitive a function is in relation to changes in its input.
The derivative of a function \( f(x) \) with respect to \( x \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
In the context of our original exercise, the function \( f(x) = 2x \sin(3x) \) includes both a linear term \( 2x \) and a trigonometrically scaled sinusoidal term \( \sin(3x) \). Finding the derivative involves both the chain and product rules, reflecting the derivative's role in capturing intricate changes of a function.
Once each component derivative is found, they are pieced together to find the full derivative \( f'(x) = 2\sin(3x) + 6x\cos(3x) \). This result indicates how the original function changes with \( x \) and integrates knowledge of fundamental calculus principles.
The derivative of a function \( f(x) \) with respect to \( x \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
In the context of our original exercise, the function \( f(x) = 2x \sin(3x) \) includes both a linear term \( 2x \) and a trigonometrically scaled sinusoidal term \( \sin(3x) \). Finding the derivative involves both the chain and product rules, reflecting the derivative's role in capturing intricate changes of a function.
Once each component derivative is found, they are pieced together to find the full derivative \( f'(x) = 2\sin(3x) + 6x\cos(3x) \). This result indicates how the original function changes with \( x \) and integrates knowledge of fundamental calculus principles.
Other exercises in this chapter
Problem 15
Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(P=50 e^{-0.6 t}\)
View solution Problem 15
Find the derivative. Assume \(a, b, c, k\) are constants. $$y=4.2 q^{2}-0.5 q+11.27$$
View solution Problem 16
Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ f(z)=\sqrt{z} e^{-z} $$
View solution Problem 16
Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(P=200 e^{0.12 t}\)
View solution