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Chapter 8: Problem 3

Find \(f^{\prime}(x)\) if \(f(x)\) is the given expression. \(x^{2} \operatorname{arcsec}\left(x^{2}\right)\)

Short Answer

Expert verified
The derivative is \( f'(x) = 2x \operatorname{arcsec}(x^2) + \frac{2}{x \sqrt{x^4 - 1}} \).

Step by step solution

01

Identify Functions Involved

The function given is a product of two separate functions. We have the first function: \( u(x) = x^2 \), and the second function: \( v(x) = \operatorname{arcsec}(x^2) \).
02

Differentiate the First Function

The derivative of \( u(x) = x^2 \) with respect to \( x \) is computed as follows: \( u'(x) = 2x \).
03

Differentiate the Second Function

The derivative of \( v(x) = \operatorname{arcsec}(x^2) \) requires the chain rule. The derivative of \( \operatorname{arcsec}(x) \) is \( \frac{1}{x\sqrt{x^2-1}} \). Thus, the derivative of \( \operatorname{arcsec}(x^2) \) is \( v'(x) = \frac{2x}{x^4 \sqrt{x^4 - 1}} \) after applying the chain rule.
04

Apply the Product Rule

The product rule states that \((uv)' = u'v + uv'\). Applying this gives:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]Substitute the expressions from previous steps:\[ f'(x) = (2x)\operatorname{arcsec}(x^2) + x^2 \left( \frac{2x}{x^4 \sqrt{x^4 - 1}} \right) \]
05

Simplify the Expression

Simplify the expression obtained in Step 4:\[ f'(x) = 2x \operatorname{arcsec}(x^2) + \frac{2x^3}{x^4 \sqrt{x^4 - 1}} = 2x \operatorname{arcsec}(x^2) + \frac{2}{x \sqrt{x^4 - 1}} \]

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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 key tool in calculus for finding derivatives of composite functions. In essence, it allows us to differentiate an outer function and an inner function simultaneously, following a specific method. If you have a composite function, say \( f(g(x)) \), the derivative is found by differentiating the outer function \( f \) and multiplying it by the derivative of the inner function \( g \.x\). This gives us the chain rule formula:
  • \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
It's important to recognize when to use the chain rule. Whenever you see a function within another function, consider if the chain rule applies. Applying it correctly makes solving otherwise complex differentiation problems manageable.
Product Rule
The product rule is used in calculus to find the derivative of a product of two functions. If you have a product of functions, \( u(x) \cdot v(x) \), its derivative cannot be simply the product of their individual derivatives. Instead, the product rule states:
  • \((uv)' = u'v + uv'\)
This rule requires you to first differentiate one function and then multiply it by the other unaltered function. After that, differentiate the second function and multiply it by the first unaltered function. Finally, add these two results together. Recognizing combinations of functions as products is critical when applying the product rule effectively during differentiation.
Arcsecant Function
The arcsecant function is the inverse of the secant function, denoted as \( \operatorname{arcsec}(x) \). Finding its derivative is often less straightforward than more common functions. The derivative formula of \( \operatorname{arcsec}(x) \) with respect to \( x \) is:
  • \( \frac{d}{dx}\left[\operatorname{arcsec}(x)\right] = \frac{1}{x\sqrt{x^2-1}} \)
This derivative arises from the unique nature of the arcsecant, involving both \( x \) and the square root of \( x^2 - 1 \). When dealing with transformations like \( \operatorname{arcsec}(x^2) \), apply the chain rule for differentiation, as seen in the example provided. This ensures the correct treatment of the internal function \( x^2 \).
Differentiation Steps
Differentiation is a process of finding the rate at which one variable changes with respect to another. When differentiating expressions involving multiple components like products or composite functions, a step-by-step approach is crucial. Here is a simplified outline:
  • Recognize the form of your function, identifying if it's a product or a composite.
  • Differentiate each component separately using appropriate rules: the product rule or the chain rule.
  • Combine the differentiated parts to form the derivative of the entire expression.
  • Simplify the result, ensuring it is presentable and readable. Simplification might involve combining like terms or reducing fractions.
Being systematic in your approach with differentiation helps prevent errors and clarifies the solution process. It's essential to understand the function's structure to apply the correct differentiation rules effectively.

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