Chapter 3: Problem 5
Write the composite function in the form \(f(g(x))\) . [Identify the inner function \(u=g(x)\) and the outer function \(y=f(u) . ]\) Then find the derivative \(d y / d x\). \(y=e^{\sqrt{x}}\)
Short Answer
Expert verified
The derivative is \( \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \).
Step by step solution
01
Identify the Inner Function
The first step is to identify the inner function in the given expression. We have the expression of the function: \( y = e^{\sqrt{x}} \). Consider the inner operation, which involves \( \sqrt{x} \). Hence, the inner function is \( g(x) = \sqrt{x} \).
02
Identify the Outer Function
Next, identify the outer function based on the expression that is applied to the result of the inner function. Here, we have \( y = e^{u} \), where \( u = g(x) = \sqrt{x} \). Hence, the outer function is \( f(u) = e^u \).
03
Express as Composite Function
Now, let's write the given function \( y = e^{\sqrt{x}} \) as a composite function: \( y = f(g(x)) = f(u) = e^{g(x)} = e^{\sqrt{x}} \). The expression for \( f(g(x)) \) confirms the composite nature of the function.
04
Find the Derivative \( \frac{dy}{dx} \)
To find \( \frac{dy}{dx} \), use the chain rule. The derivative of \( f(u) = e^u \) with respect to \( u \) is \( \frac{dy}{du} = e^u \). The derivative of \( g(x) = \sqrt{x} \) with respect to \( x \) is \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \). Applying the chain rule gives \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = e^{\sqrt{x}} \times \frac{1}{2\sqrt{x}} \).
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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, especially when working with composite functions. It allows us to differentiate functions that are composed of two or more simpler functions. Whenever you have a composite function, such as \( f(g(x)) \), you'll need the chain rule to find its derivative.
Here’s how it works:
The beauty of the chain rule lies in its ability to "chain" the derivatives together, providing a bridge between the inner and outer functions. This approach is crucial when the function is layered or when its composition involves several functions.
Here’s how it works:
- First, identify the outer function \( f \), which is applied last in the sequence of operations.
- Then, identify the inner function \( g \), which is applied first.
- Use the chain rule formula: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \), where \( y = f(g(x)) \).
The beauty of the chain rule lies in its ability to "chain" the derivatives together, providing a bridge between the inner and outer functions. This approach is crucial when the function is layered or when its composition involves several functions.
Derivative Calculation
Calculating the derivative of a composite function involves applying the chain rule we just discussed. It's a straightforward but powerful technique in differentiating complex functions.
Follow these steps for the calculation:
Follow these steps for the calculation:
- Identify the inner and outer functions of the composite.
- Differentiate the outer function with respect to the inner function. This gives you \( \frac{dy}{du} \).
- Differentiate the inner function with respect to \( x \), resulting in \( \frac{du}{dx} \).
- Multiply these derivatives together to get \( \frac{dy}{dx} \).
- The outer function \( f(u) = e^u \) gives \( \frac{dy}{du} = e^u \).
- The inner function \( g(x) = \sqrt{x} \) results in \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \).
- Use the chain rule to find \( \frac{dy}{dx} = e^{\sqrt{x}} \times \frac{1}{2\sqrt{x}} \).
Inner and Outer Functions
In composite functions, dissecting them into their inner and outer components is crucial for understanding and differentiation. Here is why knowing these components is important:
Linking these functions through composition is like putting one inside the other, creating a sequence of operations leading to the final expression. Understanding this relationship simplifies the process of finding the derivative and enhances comprehension of how the function behaves.
- The inner function \( g(x) \): This is the initial transformation applied to the input \( x \). For example, in \( y = e^{\sqrt{x}} \), \( g(x) = \sqrt{x} \) is the inner function because \( x \) undergoes square root transformation first.
- The outer function \( f(u) \): This is the subsequent transformation applied to the result of the inner function. In our earlier function, \( f(u) = e^u \) represents the outer function, as it operates on the output \( u \) from \( g(x) \).
Linking these functions through composition is like putting one inside the other, creating a sequence of operations leading to the final expression. Understanding this relationship simplifies the process of finding the derivative and enhances comprehension of how the function behaves.