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Magazine Chapter 3: Problem 42
In Exercises \(1-56,\) find the derivatives. Assume that \(a\) and \(b\) are constants. $$h(x)=2^{e^{3 x}}$$
Short Answer
Expert verified
The derivative is \( h'(x) = 3 \cdot 2^{e^{3x}} \cdot e^{3x} \ln(2) \).
Step by step solution
01
Understanding the Function
The given function is \( h(x) = 2^{e^{3x}} \). This is an exponential function where the base is 2 and the exponent is an exponential expression \( e^{3x} \). To differentiate it, we'll use the chain rule, as it involves a compound function.
02
Differentiate the Outer Function
The outer function is \( 2^u \) where \( u = e^{3x} \). The derivative of \( 2^u \) with respect to \( u \) is \( 2^u \ln(2) \) according to the chain rule for differentiation of exponential functions with bases other than \( e \).
03
Differentiate the Inner Function
Now, differentiate the inner function \( u = e^{3x} \). Its derivative with respect to \( x \) is \( \frac{d}{dx}(e^{3x}) = 3e^{3x} \), due to the chain rule where the derivative of \( e^v \) with respect to \( v \) is \( e^v \), multiplied by the derivative of \( v = 3x \), which is 3.
04
Apply the Chain Rule
Combine the derivatives from Step 2 and Step 3 using the chain rule. The chain rule states that \( \frac{dh}{dx} = \frac{dh}{du} \cdot \frac{du}{dx} \). Substituting the expressions derived, this results in:\[ h'(x) = 2^{e^{3x}} \ln(2) \cdot 3e^{3x} \]
05
Simplify the Expression
Lastly, simplify the expression for the derivative:\[ h'(x) = 3 \cdot 2^{e^{3x}} \cdot e^{3x} \ln(2) \]No further simplification is needed, so this is the derivative of \( h(x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus that plays a crucial role in differentiating composite functions. When you have a function within another function, like in our case with the given exercise, the chain rule helps us find the derivative.
Consider the function hierarchy: the outer function and the inner function. In our exercise, the outer function is expressed as \(2^u\), and the inner function is \(u = e^{3x}\).
Here's how you apply the chain rule:
Consider the function hierarchy: the outer function and the inner function. In our exercise, the outer function is expressed as \(2^u\), and the inner function is \(u = e^{3x}\).
Here's how you apply the chain rule:
- First, find the derivative of the outer function with regard to its inner function, so for \(2^u\), you differentiate it as \(2^u \ln(2)\).
- Next, find the derivative of the inner function with respect to \(x\), which is \(3e^{3x}\) for \(e^{3x}\).
- Finally, multiply these derivatives together: \(\frac{dh}{dx} = \frac{dh}{du} \cdot \frac{du}{dx}\).
Exponential Function
Exponential functions are a class of functions that feature a constant base raised to a variable exponent. In this exercise, the function \(h(x) = 2^{e^{3x}}\) is an example of an exponential function with compound exponents.
Key properties of exponential functions include:
Key properties of exponential functions include:
- The base here is constant, i.e., \(2\), which indicates that it identifies a growth or decay pattern.
- The exponent \(e^{3x}\) introduces another level of complexity, making the differentiation process more engaging.
- A remarkable feature of exponential functions is that their derivatives mirror the structure of the original function, with a constant scaling, depending on the base.
Differentiation
Differentiation refers to the process in calculus where we calculate the derivative of a function. This tells us how the function changes as the input variable changes, and is essential in understanding the behavior of mathematical functions.
Steps for differentiation in compound functions like \(h(x) = 2^{e^{3x}}\) involve:
Steps for differentiation in compound functions like \(h(x) = 2^{e^{3x}}\) involve:
- Identifying the type of functions you are working with, here emphasizing the exponential nature.
- Utilizing the appropriate differentiation rules, such as the chain rule for complex expressions.
- Combining partial derivatives to find a complete derivative expression.
- We began by differentiating \(2^u\) with respect to \(u\), resulting in \(2^u \ln(2)\).
- We then derived \(e^{3x}\) concerning \(x\), which gave us \(3e^{3x}\).
