Problem 24
Question
Calculate the derivative with respect to \(x\) of the given expression. \(3^{x} \cdot \exp \left(x^{2}\right)\)
Step-by-Step Solution
Verified Answer
The derivative of the expression is \(3^x \cdot \exp(x^2) \cdot (\ln(3) + 2x)\).
1Step 1: Identify the components of the expression
The expression we need to differentiate is a product of two functions: \(f(x) = 3^x\) and \(g(x) = \exp(x^2)\). According to the product rule for derivatives, the derivative of a product \(f(x) \cdot g(x)\) is given by \(f'(x) \cdot g(x) + f(x) \cdot g'(x)\). We need to differentiate both \(f(x)\) and \(g(x)\).
2Step 2: Differentiate each component
First, find the derivative of \(f(x) = 3^x\):\(f'(x) = 3^x \cdot \ln(3)\), using the formula for the derivative of an exponential function with base \(a\), which is \(a^x \cdot \ln(a)\).Next, find the derivative of \(g(x) = \exp(x^2)\):\(g'(x) = \exp(x^2) \cdot 2x\), using the chain rule where the outer function is the exponential and the inner function is \(x^2\).
3Step 3: Apply the product rule
Substitute the derivatives found in Step 2 into the product rule formula: \(\frac{d}{dx}[3^x \cdot \exp(x^2)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)\).So, we have:\(\frac{d}{dx}[3^x \cdot \exp(x^2)] = 3^x \cdot \ln(3) \cdot \exp(x^2) + 3^x \cdot \exp(x^2) \cdot 2x\).
4Step 4: Simplify the expression
Factor out the common terms from both parts of the expression found in Step 3:\(\frac{d}{dx}[3^x \cdot \exp(x^2)] = 3^x \cdot \exp(x^2) \cdot (\ln(3) + 2x)\).This is the simplified form of the derivative.
Key Concepts
Understanding the Product RuleThe Chain Rule SimplifiedWorking with Exponential Functions
Understanding the Product Rule
When you are faced with a function that is the product of two or more functions, the product rule is your go-to strategy for finding the derivative. This rule is essential because it allows us to break down complex expressions into more manageable pieces.
To apply the product rule, consider a product of two functions:
In our exercise:
To apply the product rule, consider a product of two functions:
- Function one: \( f(x) \).
- Function two: \( g(x) \).
In our exercise:
- \( f(x) = 3^x \), and \( g(x) = \exp(x^2) \).
The Chain Rule Simplified
The chain rule is fundamental when differentiating compositions of functions, especially with nested functions. It provides a systematic way to handle equations where one function is plugged into another.
The basic idea is to determine the derivative of the outer function and multiply it by the derivative of the inner function.
For a composite function \( h(x) = f(g(x)) \), the chain rule says: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \] In the exercise, the function \( g(x) = \exp(x^2) \) presents an excellent example. Here, the outer function is \( \exp(u) \) where \( u = x^2 \), and the inner function is \( x^2 \).
We differentiate the outer function \( f(u) = \exp(u) \) to get \( \exp(x^2) \), and the inner function \( x^2 \) to get \( 2x \). Multiply these results for our chain rule and we see how efficiently this rule handles layered functions. It’s a powerful tool in a derivative toolbox, helping us go deeper into understanding functions.
The basic idea is to determine the derivative of the outer function and multiply it by the derivative of the inner function.
For a composite function \( h(x) = f(g(x)) \), the chain rule says: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \] In the exercise, the function \( g(x) = \exp(x^2) \) presents an excellent example. Here, the outer function is \( \exp(u) \) where \( u = x^2 \), and the inner function is \( x^2 \).
We differentiate the outer function \( f(u) = \exp(u) \) to get \( \exp(x^2) \), and the inner function \( x^2 \) to get \( 2x \). Multiply these results for our chain rule and we see how efficiently this rule handles layered functions. It’s a powerful tool in a derivative toolbox, helping us go deeper into understanding functions.
Working with Exponential Functions
Exponential functions, where the variable appears in the exponent, have unique properties that differentiate them from basic polynomial or trigonometric functions. In our exercise, \( 3^x \) and \( \exp(x^2) \) are exponential.
For exponential functions of the form \( a^x \), the derivative involves logarithms. The derivative formula is: \[\frac{d}{dx}[a^x] = a^x \cdot \ln(a) \]This is due to the constant growth rate characteristic of exponential functions.
The natural exponential function, \( \exp(x) \) or \( e^x \), is special. Its derivative is itself: \[\frac{d}{dx}[e^x] = e^x \]In our problem, \( g(x) = \exp(x^2) \) utilized this property in conjunction with the chain rule for a swift determination of its derivative.
Exponential functions are invaluable in modeling real-world phenomena, representing growth or decay, making them key in fields ranging from biology to economics. Understanding how to derive them is a crucial mathematical skill.
For exponential functions of the form \( a^x \), the derivative involves logarithms. The derivative formula is: \[\frac{d}{dx}[a^x] = a^x \cdot \ln(a) \]This is due to the constant growth rate characteristic of exponential functions.
The natural exponential function, \( \exp(x) \) or \( e^x \), is special. Its derivative is itself: \[\frac{d}{dx}[e^x] = e^x \]In our problem, \( g(x) = \exp(x^2) \) utilized this property in conjunction with the chain rule for a swift determination of its derivative.
Exponential functions are invaluable in modeling real-world phenomena, representing growth or decay, making them key in fields ranging from biology to economics. Understanding how to derive them is a crucial mathematical skill.
Other exercises in this chapter
Problem 24
Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{\pi}^{3 \pi} \pi d x $$
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Compute the average value of \(f\) over \([a, b]\). \(f(x)=e^{x} \quad a=0, b=2\)
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In each of Exercises \(21-28,\) calculate the derivative of \(F(x)\) with respect to \(x\). $$ F(x)=\int_{0}^{x} \tan ^{2}(t) d t $$
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In Exercises \(23-26,\) use an identity to simplify the sum. $$ \sum_{j=0}^{5} \frac{2^{2 j-1}}{3^{j+1}} $$
View solution