Problem 72
Question
Find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=\cos ^{2}\left(\sqrt{2 x^{2}+3}\right) $$
Step-by-Step Solution
Verified Answer
The derivative is \(-\frac{4x \cos(\sqrt{2x^2 + 3}) \sin(\sqrt{2x^2 + 3})}{\sqrt{2x^2 + 3}}\)."
1Step 1: Identify the outer function and its derivative
The outer function is \( u^2 \) where \( u = \cos \left(\sqrt{2x^2 + 3}\right) \). The derivative of \( u^2 \) with respect to \( u \) is \( 2u \).
2Step 2: Apply the chain rule to the outer function
Using the chain rule, the derivative of \( f(x) \) is \( 2u \cdot \frac{du}{dx} \), where \( u = \cos \left(\sqrt{2x^2 + 3}\right) \). Simplifying gives \( 2 \cos\left(\sqrt{2x^2 + 3}\right) \cdot \frac{d}{dx} \left(\cos \left(\sqrt{2x^2 + 3}\right)\right) \).
3Step 3: Differentiate the inner function (Cosine function)
The inner function inside the cosine is \( \cos \left(v\right) \) where \( v = \sqrt{2x^2 + 3} \). The derivative of \( \cos(v) \) is \( -\sin(v) \cdot \frac{dv}{dx} \).
4Step 4: Differentiate the innermost function (Square root function)
Here, \( v = \sqrt{2x^2 + 3} = (2x^2 + 3)^{1/2} \). Differentiate to get \( \frac{dv}{dx} = \frac{1}{2\sqrt{2x^2 + 3}} \cdot \frac{d}{dx}(2x^2 + 3) \).
5Step 5: Differentiate the polynomials inside the square root
The derivative of \( 2x^2 + 3 \) is \( 4x \). Thus, \( \frac{dv}{dx} = \frac{4x}{2\sqrt{2x^2 + 3}} = \frac{2x}{\sqrt{2x^2 + 3}} \).
6Step 6: Combine all derivatives using the chain rule
Substitute back into our chain rule expressions: \[ f'(x) = 2 \cos\left(\sqrt{2x^2 + 3}\right) \cdot (-\sin\left(\sqrt{2x^2 + 3}\right)) \cdot \frac{2x}{\sqrt{2x^2 + 3}} \]. So, \[ f'(x) = -4x \cos\left(\sqrt{2x^2 + 3}\right) \sin\left(\sqrt{2x^2 + 3}\right) / \sqrt{2x^2 + 3}\].
Key Concepts
Chain RuleTrigonometric DifferentiationComposite Functions
Chain Rule
The chain rule is a fundamental concept in derivative calculus, particularly useful when dealing with composite functions. It allows us to find the derivative of a composite function by breaking it down into simpler parts. Essentially, if we have a function that is composed of another function, say if we have a function of a function, we use the chain rule to differentiate it.
In a more mathematical sense, if we have a composite function like \( f(g(x)) \), the derivative \( f'(x) \) can be found using \( f'(g(x)) \cdot g'(x) \). We begin by differentiating the outer function and then multiply it by the derivative of the inner function.
For example, in our exercise, we have a function \( f(x) = \cos^2(\sqrt{2x^2 + 3}) \), where \( \cos^2(u) \) is our outer function. By identifying our inner functions step by step, we are able to apply the chain rule sequentially. By treating each composite layer as its own function, we are able to find the derivative efficiently.
In a more mathematical sense, if we have a composite function like \( f(g(x)) \), the derivative \( f'(x) \) can be found using \( f'(g(x)) \cdot g'(x) \). We begin by differentiating the outer function and then multiply it by the derivative of the inner function.
For example, in our exercise, we have a function \( f(x) = \cos^2(\sqrt{2x^2 + 3}) \), where \( \cos^2(u) \) is our outer function. By identifying our inner functions step by step, we are able to apply the chain rule sequentially. By treating each composite layer as its own function, we are able to find the derivative efficiently.
Trigonometric Differentiation
Trigonometric differentiation is a process used when dealing with functions that include trigonometric elements, such as sine, cosine, and tangent. Differentiating these functions follows specific rules that are derived from the unit circle definitions and the properties of trigonometric functions.
For instance, the derivative of \( \cos(x) \) is \( -\sin(x) \). This is crucial when we apply the chain rule to composite functions, as shown in our problem. We deal with functions that involve \( \cos \) and use this differentiation rule to translate changes in our angle into rate changes in our function value.
In the exercise we are discussing, differentiation of the cosine function which is wrapped around another function—in this case, \( \cos(\sqrt{2x^2 + 3}) \)—requires an understanding of both the chain rule and trigonometric differentiation. Recognizing the derivative forms the basis from which we compute more complex derivatives by chain rule application.
For instance, the derivative of \( \cos(x) \) is \( -\sin(x) \). This is crucial when we apply the chain rule to composite functions, as shown in our problem. We deal with functions that involve \( \cos \) and use this differentiation rule to translate changes in our angle into rate changes in our function value.
In the exercise we are discussing, differentiation of the cosine function which is wrapped around another function—in this case, \( \cos(\sqrt{2x^2 + 3}) \)—requires an understanding of both the chain rule and trigonometric differentiation. Recognizing the derivative forms the basis from which we compute more complex derivatives by chain rule application.
Composite Functions
Understanding composite functions is key to mastering calculus problems that involve the chain rule and trigonometric differentiation. A composite function is simply a function that takes another function as its input, written as \( f(g(x)) \). It’s like having an operation within another operation. For instance, in our exercise, the outer function is \( \cos^2(u) \) and the inner function is \( \sqrt{2x^2 + 3} \).
Working with composite functions involves unwinding each layer and differentiating in parts, often from the outermost to the innermost layer. The derivative of a composite function is determined by the derivative of the outer function multiplied by the derivative of the inner function.
Through practice, recognizing composite functions and applying the chain rule to them becomes intuitive, allowing you to tackle more complex calculus problems with confidence. Each layer adds a piece to the puzzle, collectively leading to the final derivative form, as demonstrated in the given example.
Working with composite functions involves unwinding each layer and differentiating in parts, often from the outermost to the innermost layer. The derivative of a composite function is determined by the derivative of the outer function multiplied by the derivative of the inner function.
Through practice, recognizing composite functions and applying the chain rule to them becomes intuitive, allowing you to tackle more complex calculus problems with confidence. Each layer adds a piece to the puzzle, collectively leading to the final derivative form, as demonstrated in the given example.
Other exercises in this chapter
Problem 71
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