Problem 40
Question
Find the indicated derivative. \(D_{x}\left(3^{2 x^{2}-3 x}\right)\)
Step-by-Step Solution
Verified Answer
The derivative is \( 3^{2x^2 - 3x} \cdot \ln(3) \cdot (4x - 3) \).
1Step 1: Identify the Function
The given function is an exponential function: \( f(x) = 3^{2x^2 - 3x} \). We need to find the derivative of this function with respect to \( x \).
2Step 2: Rewrite the Exponential Function using the Natural Exponent
Recall that any exponential function \( a^x \) can be rewritten using the natural number \( e \) as \( e^{x \ln(a)} \). Apply this to the given function: \( f(x) = e^{(2x^2 - 3x) \ln(3)} \).
3Step 3: Differentiate using the Chain Rule
To differentiate \( f(x) = e^{u} \) where \( u = (2x^2 - 3x) \ln(3) \), we use the chain rule: \( \frac{d}{dx}[e^u] = e^u \cdot \frac{du}{dx} \).
4Step 4: Find the Derivative of u with respect to x
The function \( u = (2x^2 - 3x) \ln(3) \) is a product. Differentiate \( u \) as \( \frac{du}{dx} = \ln(3) \cdot (4x - 3) \).
5Step 5: Combine Results for the Derivative
Plug \( e^u = e^{(2x^2 - 3x) \ln(3)} = 3^{2x^2 - 3x} \) and \( \frac{du}{dx} = \ln(3) \cdot (4x - 3) \) into the chain rule result: \( D_x(3^{2x^2 - 3x}) = 3^{2x^2 - 3x} \cdot \ln(3) \cdot (4x - 3) \).
Key Concepts
Chain RuleExponential FunctionsCalculus
Chain Rule
The chain rule is an essential concept in calculus, especially when dealing with composite functions. Imagine function g(x) is embedded within another function f(u). To find the derivative of the entire composition, we use the chain rule. The chain rule formula is:
For example, if you have a situation like in the exercise where the function \( f(x) = e^u \) with \( u = (2x^2 - 3x) \ln(3) \), the chain rule suggests that we differentiate \( e^u \) treating \( u \) as constant, then multiply by \( \frac{du}{dx} \), the derivative of \( u \).
This is helpful when differentiating complex functions expressed in terms of simpler functions, like exponential functions in our task.
- \( \frac{d}{dx}\big[ f(g(x)) \big] = f'(g(x)) \cdot g'(x) \)
For example, if you have a situation like in the exercise where the function \( f(x) = e^u \) with \( u = (2x^2 - 3x) \ln(3) \), the chain rule suggests that we differentiate \( e^u \) treating \( u \) as constant, then multiply by \( \frac{du}{dx} \), the derivative of \( u \).
This is helpful when differentiating complex functions expressed in terms of simpler functions, like exponential functions in our task.
Exponential Functions
Exponential functions represent a key area of calculus due to their unique properties, particularly their rate of growth. A general form of such a function is \( a^x \), where \( a \) is a constant, and \( x \) is the variable exponent.
The exponential function is notable because its derivative has a specific property: the derivative of \( e^x \) is \( e^x \) itself. For other bases, such as \( 3^x \), the derivative involves the natural logarithm of the base: \( \frac{d}{dx}[a^x] = a^x \ln(a) \).
In exercises like ours, after rewriting the base \( 3 \) using the natural logarithm, we differentiate the exponential function as if it were \( e^u \), where \( u \) incorporates all variable-dependent parts. This requires the use of the chain rule to accurately compute the derivative.
The exponential function is notable because its derivative has a specific property: the derivative of \( e^x \) is \( e^x \) itself. For other bases, such as \( 3^x \), the derivative involves the natural logarithm of the base: \( \frac{d}{dx}[a^x] = a^x \ln(a) \).
In exercises like ours, after rewriting the base \( 3 \) using the natural logarithm, we differentiate the exponential function as if it were \( e^u \), where \( u \) incorporates all variable-dependent parts. This requires the use of the chain rule to accurately compute the derivative.
Calculus
Calculus is the branch of mathematics that explores changes, represented through derivatives and integrals. At its core, calculus deals with understanding rates of change and the accumulation of quantities.
In the context of derivatives, calculus provides the tools to determine how a function changes over an interval. This is crucial for solving problems involving motion, growth, and areas under curves, among others.
Exercises like these expand your problem-solving toolkit, refining your ability to apply these principles to both real-world and theoretical situations.
In the context of derivatives, calculus provides the tools to determine how a function changes over an interval. This is crucial for solving problems involving motion, growth, and areas under curves, among others.
- Derivatives: These describe the rate at which a function is changing at any point. Calculating a derivative means finding another function that gives the slope of the tangent line at any point on the graph.
- Integrals: In contrast, integrals explore the area under a curve, accumulated over an interval of the function.
Exercises like these expand your problem-solving toolkit, refining your ability to apply these principles to both real-world and theoretical situations.
Other exercises in this chapter
Problem 39
In Problems 33-40, apply the Chain Rule more than once to find the indicated derivative. $$ \frac{d}{d x}\\{\sin [\cos (\sin 2 x)]\\} $$
View solution Problem 40
Use implicit differentiation twice to find \(y^{\prime \prime}\) at \((3,4)\) if \(x^{2}+y^{2}=25\).
View solution Problem 40
$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{5 x-4}{3 x^{2}+1} $$
View solution Problem 40
In Problems 33-40, apply the Chain Rule more than once to find the indicated derivative. $$ \frac{d}{d t}\left\\{\cos ^{2}[\cos (\cos t)]\right\\} $$
View solution